ISI CMI IMPORTANT AND COMPLICATED QUESTION WITH SOLUTION BY SOURAV SIR'S CLASSES 9836793076

1

SOME IMPORTANT

AND COMPLICATED

QUESTIONS FOR

CMI ENTRANCE

EXAMINATION

BY

SOURAV SIR’S CLASSES

KOLKATA & NEW DELHI

9836793076

2

Question

A system of equations has 3 equations

3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + kz = 6.

If the system of equations has no solution, find k.

Explanation

This is a very interesting question. If we can generate one of the equations from the other

two, we can then say that the system has infinite solutions. If we can generate one of the

equations from the other two, but with the constant part alone being different, this would

be akin to having parallel lines when we are dealing with 2 variables and that would

result in the system having no solutions. So, lets look for that.

So, we need to find some way where a(3x + 4y + 5z) + b(4x + 5y + 3z) = 2x + 3y + kz.

We need to find k. In other words, we need to find a, b such that a(3x + 4y) + b(4x + 5y )

= 2x + 3y . Then said a, b would give us k.

Or, we are effectively solving for

3a + 4b = 2

4a + 5b = 3

Subtracting one from the other, we get a + b = 1. 3a + 3b = 3. Or, b = -1. a = 2.

Now, k = 5a + 3b = 10 -3 = 7.

If k = 7, this system of equations would result in no solutions.

If k were 7, the system of equations should be 3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and

2x + 3y + 7z = 6.

First equation * 2 - second equation would give us the equation 2x + 3y + 7z = 8.

Now, 2x + 3y + 7z cannot be 6 and 8 at the same time. So, this system of equations has

no solution.

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Yeah, the more mechanical, rather deceptively straight-forward-looking determinant

method is also there. But where is the joy in that.

1. a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one of ab,

bc and ca is odd. abc is a multiple of 4. The arithmetic mean of a and b is an

integer and so is the arithmetic mean of a, b and c. How many such triplets are

possible (unordered triplets)

Exactly one of ab, bc and ca is odd => Two are odd and one is even

abc is a multiple of 4 => the even number is a multiple of 4

The arithmetic mean of a and b is an integer => a and b are odd

and so is the arithmetic mean of a, b and c. => a+ b + c is a multiple of 3

c can be 4 or 8.

c = 4; a, b can be 3, 5 or 5, 9

c = 8; a, b can be 3, 7 or 7, 9

Four triplets are possible

2. A seven-digit number comprises of only 2's and 3's. How many of these are

multiples of 12?

Number should be a multiple of 3 and 4. So, the sum of the digits should be a

multiple of 3. WE can either have all seven digits as 3, or have three 2's and four

3's, or six 2's and a 3. (The number of 2's should be a multiple of 3).

For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us

combine these two.

All seven 3's - No possibility

4

Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some

order. No of possibilities = 5!/3!*2! = 10

Six 2's and one 3 - The first 5 digits should all be 2's. So, there is only one number

2222232

3. How many factors of 2^5 * 3^6 * 5^2 are perfect squares?

Any factor of this number should be of the form 2^a * 3^b * 5^c. For the factor to

be a perfect square a,b,c have to be even. a can take values 0, 2, 4. b can take

values 0,2, 4, 6 and c can take values 0,2. Total number of perfect squares = 3 * 4 *

2 = 24

4. How many factors of 2^4 * 5^3 * 7^4 are odd numbers?


be an odd number, a should be 0. b can take values 0,1, 2,3, and c can take values

0, 1, 2,3, 4. Total number of odd factors = 4 * 5 = 20

5. A number when divided by 18 leaves a remainder 7. The same number when

divided by 12 leaves a remainder n. How many values can n take?

Number can be 7, 25, 43, 61, 79.

Remainders when divided by 12 are 7 and 1.

Question A system of equations has 3 equations

3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and 2x + 3y + kz = 6.

If the system of equations has no solution, find k.

Explanation This is a very interesting question. If we can generate one of the equations from the other

two, we can then say that the system has infinite solutions. If we can generate one of the

equations from the other two, but with the constant part alone being different, this would

be akin to having parallel lines when we are dealing with 2 variables and that would

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result in the system having no solutions. So, lets look for that.

So, we need to find some way where a(3x + 4y + 5z) + b(4x + 5y + 3z) = 2x + 3y + kz.

We need to find k. In other words, we need to find a, b such that a(3x + 4y) + b(4x + 5y )

= 2x + 3y . Then said a, b would give us k.

Or, we are effectively solving for

3a + 4b = 2

4a + 5b = 3

Subtracting one from the other, we get a + b = 1. 3a + 3b = 3. Or, b = -1. a = 2.

Now, k = 5a + 3b = 10 -3 = 7.

If k = 7, this system of equations would result in no solutions.

If k were 7, the system of equations should be 3x + 4y + 5z = 11, 4x + 5y + 3z = 14 and

2x + 3y + 7z = 6.

First equation * 2 - second equation would give us the equation 2x + 3y + 7z = 8.

Now, 2x + 3y + 7z cannot be 6 and 8 at the same time. So, this system of equations has

no solution.

Yeah, the more mechanical, rather deceptively straight-forward-looking determinant

method is also there. But where is the joy in that.

Question Product of the distinct digits of a natural number is 60. How many such numbers are

possible?

Explanation 60 = 2^2 * 3 * 5

60 cannot be written as a product of two single digit numbers. SO, the number in question

should either have 3 or more digits.

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Three-digit numbers

The digits could be 345 or 265

Digits being 345 - there are 3! such numbers

There are six numbers for each of these outlines. So, there are 3! + 3! = 12 three-digit

numbers

Four-digit numbers

The digits could be 1345, 1265 or 2235.



Digits being 2235 - this is not possible as digits have to be distinct.

So, there are totally 24 + 24 four-digit numbers possible. 48 four-digit numbers.

Total number of numbers = 12 + 48 = 60.

John has chocolates of types A and B in the ratio 3 : 7, while Mike has chocolates of

types B and C in the ratio 5 : 4, Ram has chocolates of types C and A in the ratio 3 : 5. If

there are more chocolates of type C than of type B, and more of type B than of type A,

what is the minimum possible number of chocolates overall?

Explanation

Again, big thanks to Mukund Sukumar for excellent solution.

Let john have 3x chocolates of type A and 7x of type B

Let Mike have 5y chocolates of type B and 4y of type C

Let john have 3z chocolates of type C and 5z of type A

So in total A=3x+5z ; B=7x+5y ; C=4y+3z

Since C>B we get solving y<3z-7x --->(1)

Since B>A we get solving 5y>5z-4x ---> (2)

What gets inferred from above 2 statements is z>=3. so when x=1,z=3, we get only y=1

as choice, for which second condition doesnt satisfy.

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So, when x=1,z=4, we get y<5 from first condition and when y > 3.2 from second

condition. so which gives choice the only y=4.

Hence x=1,y=4 and z=4 works and is the best possible answer.

For these values, we get A=23,B=27,C=28.

Minimum possible number of chocolates overall is 76

A wonderful, but very tough question from Permutation and Combinations.

In how many ways, can we rearrange the word MONSOON such that no two adjacent

positions are taken by the same letter? (Tough one. Tougher than what we will see in

CAT)

Explanation

First up, lets get the facts out of the way – The three O’s need to be kept apart, then the 2

N’s.

Let us focus on the three O’s.

We can place the three O’s in some blanks around the other letters. Or, three O’s can be

placed in 3 slots out of the 5 in __M__N__S__N__ . This can be done in 5C3, or 10 ways.

Or, the O’s can be in slots {1, 3, 5} or {1, 3, 6} or {1, 3, 7} or {1, 4, 6} or {1, 4, 7} or {1,

5, 7} or {2, 4, 6} or {2, 4, 7} or {2, 5, 7} or {3, 5, 7} - Whew.

Now, for each of these arrangements, there are 4!/2! = 12 arrangements for the other 4

letters. But the one thing we need to keep in mind now is the fact that 2 N’s could be

adjacent in these arrangements. We will need to eliminate these.

O’s in slots {1, 3, 5} or O__O__O__ __ - Ns could be in the last two slots. There are 2! =

2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 3, 6} or O__O__ __O__ - Ns could be two slots 4 and 5. There are 2! = 2

words like this. So, number of words that we need to count = 12 – 2 = 10

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O’s in slots {1, 3, 7} or O__O__ __ __O - Ns could be in the slots {4, 5} or {5, 6}. There

are totally 4 words that we need to subtract. So, number of words that we need to count =

12 – 4 = 8

O’s in slots {1, 4, 6} or O__ __O__O__ - Ns could be in the slots {2, 3}. There are 2! =

2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 4, 7} or O__ __O__ __O - Ns could be in slots {2, 3} or {5, 6}. There are

totally 4 words that we need to subtract. So, number of words that we need to count = 12

– 4 = 8

O’s in slots {1, 5, 7} or O__ __ __O__ O - Ns could be in the slots {2, 3} or {3, 4}.

There are totally 4 words that we need to subtract. So, number of words that we need to

count = 12 – 4 = 8

O’s in slots {2, 4, 6} or __O__O__O__ - Tehre are no possible slots for N. So, we count

all 12 words on this list.

O’s in slots {2, 4, 7} or __O__O__ __ O - Ns could be in slots {5, 6}. There are 2! = 2

words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {2, 5, 7} or __O__ __O__ O - Ns could be in slots {3, 4}. There are 2! = 2

words like this. So, number of words that we need to count = 12 – 2 = 10.

O’s in slots {3, 5, 7} or __ __ O__O__O - Ns could be in the first two slots. There are 2!

= 2 words like this. So, number of words that we need to count = 12 – 2 = 10

Total number of words = 10 + 10 + 8 + 10 + 8 + 8 + 12 + 10 + 10 + 10 = 96.

Phew!.

There is a far more elegant method for accounting for the words where the 2 N’s appear

together. This one came from Mukund Sukumar.

We need to account for the number of possibilities of N,N being together. So to subtract

that part, consider 'NN' being together as one letter and place O's. In 3 out of the 4 slots in

_M_NN_S_

The O’s can be selected in 4C3 ways. The MNNS can be rearranged in 3! Ways if the N’s

have to appear together.

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Or, we get 4C3 * 3! = 24 ways. So, we have 120-24 = 96 ways totally.

Questions

x^2 - 17x + |p| = 0 has integer solutions. How many values can p take?

Explanation

To begin with, if we assume roots to be a and b, sum of the roots is 17 and product of the

roots is |p|. Product of the roots is positive and so is the sum of the two roots.

So, both roots need to be positive.

So, we are effectively solving for

Number of positive integer solutions for a + b = 17.

We could have (1,16), (2, 15), (3, 14)......(8, 9) - There are 8 sets of pairs of roots. Each of

these will yield a different product.

So, |p| can take 8 different values. Or, p can take 16 different values.

is that it? Or, are we missing something? Can p be zero? What are the roots of x^2 - 17x

= 0. This equation also has integer solutions.

So, p can also be zero.

So, number of possible values of p = 16 + 1 = 17.

Wonderful question - chiefly because there are two really good wrong answers one can

get. 8 and 16. So, pay attention to detail. No point telling yourself "Just missed, I just

didnt think of zero. I deserve this mark". Being just wrong, will earn us a -1 instead of the

honourable 0.

Question

How many integer solutions exist for the equation x2 - 8|x| - 48 =0?

10

Explanation

One approach is to solve when x > 0 and then solve for x < 0. However, there is a slightly

simpler method.

Note that x2 is the same as |x|2, so we can treat the equation as a quadratic in |x|.

Or, |x|2 – 8|x| - 48 = 0

(|x| - 12|) (|x| + 4) = 0

|x| could be 12. |x| cannot be -4.

If |x| could be 12, x can be -12 and 12.

Question

There are n pipes that fill a tank. Pipe 1 can fill the tank in 2 hours, Pipe 2 in 3 hours,

Pipe 3 in 4 hours and so on. Pipe 1 is kept open for 1 hour, pipe 2 for 1 hour, then pipe 3

and so on. In how many hours will the tank get filled completely?

Explanation

Almost all questions can be approached well by asking the question "What happens in 1

hour" (or 1 day, or 1 minutes)

So, let us start with that

In 1 hour, pipe 1 fills 1/2 of the tank. So, in the first hour, the tank will not be filled

In 1 hour, pipe 2 fills 1/3 of the tank. So, in two hours we would have filled 1/2 + 1/3 of

the tank, or 5/6 of the tank. So, by the end of the second hour, the tank would still not be

filled.

Let us move to the third hour. In 1 hour, pipe 3 fills 1/4 of the tank. So, by the end of the

third hour, we should have filled 5/6 + 1/4 = 13/12 of the tank.

Oops, one cannot fill 13/12 of a tank. What this tells us is that the tank gets filled in the

3rd hour.

When exactly during the third hour?

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At the beginning of the third hour, we still have 1/6 of the tank still to fill. Pipe 3 can fill

at the rate of 1/4 of the tank per hour. Or, pipe 3 will take 2/3 hours to fill the tank.

Or, the tank will be filled in 2 hours and 40 minutes.

This is an example of a sequence that is a harmonic progression. The formulae for HP are

needlessly confusing. So, simple step-by-step approach works best.

Question

How many functions can be defined from Set A -- {1, 2, 3, 4} to Set B = {a, b, c, d} that

are neither one-one nor onto?

Explanation

To start with, if you do not know the meaning of one-one or onto, look these up. For

good measure know the meanings of the terms surjective, injective etc also. CAT tests

these terms.

Let us start by answering a far simpler question. How many functions are possible from

Set A to Set B. This is equal to 4^4 = 256.

Note that any function from Set A to Set B that is one-one will also be onto and vice

versa. How? Why? - Think about this. Remember, tutors can only take the horse to the

pond. :-)

So, we need to subtract only those functions that are one-one AND onto. Or, effectively

we have to eliminate those functions where {1, 2, 3, 4} are mapped to {a, b, c, d} such

that each is mapped to a different element. This is effectively same as the number of

ways of rearranging 4 elements. Or, number of ways of doing this is 4! .

So, total number of functions that are neither one-one nor onto = 256 - 24 = 232.

Question

If we list all the words that can be formed by rearranging the letters of the word

SLEEPLESS in alphabetical order, what would be the rank of SLEEPLESS?

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Explanation

First let us think about the number of possible rearrangements. SLEEPLESS can be

rearranged in 9!/ (2! 3! 3!) = 5040.

Now, if rearrange these, we would have words starting with E, L, P and S.

Now, SLEEPLESS starts with S. So, let us think about how many words start with S.

Number of words starting with S = 8!/ (2! 2! 3!) = 1680.

So, there would have been 5040 - 1680 words that have gone by before the first word

starting with S. Or, 3360 words start with E, L or P.

The first word with S is SEEELLPSS. This would have a rank of 3361.

Now, let us go step by step.

Words starting with SE____ - Number of such words = 7!/(2!*2!*2!) = 630 words

Next we have words starting with SL, but our word also starts with SL, so let us go

deeper

Words starting with SLE would be the next step, but our word starts with SLE as

well.SO, let us go one further step deeper

Words starting with SLEE. The first such word would be SLEEEPLSS

So, let us think about words starting with SLEEE - there would be 4!/(2!) words like this

= 12 words like this

words starting with SLEEL - there would be 4!/(2!) words like this = 12 words like this

So, we have accounted for 3360 + 630 + 12 + 12 = 4014 words thus far.

Now, on to words starting with SLEEP - there are again 4!/(2!) words like this = 12

words like this

Our word is within these 12 words

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Words starting with SLEEPE - there are 3!/2! words like this, or 3 words like this

Our word comes in the next bunch.

Words starting with SLEEPL - SLEEPLESS is the first such word. Or, the rank fo

SLEEPLESS = 4014 + 3 + 1 = 4018.

A very good question to get lots of practice on letters rearrangement. However, it is

unlikely that you will face a question this time-consuming in CAT. Wonderful question

to practice though.

Question

A 3-digit Armstrong number is a three-digit number where the number is equal to the

sum of the cubes of the three digits. Give a few Armstrong numbers.

Explanation

To start with, be clear that CAT does not ask questions like these. You do not need to

know what Armstrong number are; nor do you need to know how to get these to crack

these exam. This is just a fun question to do some trial and error with.

The 3-digit Armstrong numbers are 153, 370, 371 and 407, obtained mostly by trial and

error (although a lil scientifically)

Question

How many trailing zeroes will be present in the base 12 representation of 55!?

Explanation

The question can be restated as follows - "What is the highest power of 12 that divides

55!"

Now, 12 = 2^2 * 3. So, a number that is a multiple of 2^2 and 3 will be a multiple of 12.

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So, in order to find the highest power of 12 that divides 55!, we need to look at the

highest powers of 2 and 3 that divide 55!.

Successive division by 2 will give us the highest power of 2 that divides 55!:

55/2 = 27, 27/2 = 13, 13/2 = 6, 6/2 = 3, 3/2 = 1.

Highest power of 2 that divides 55! = 27 + 13 + 6 + 3 + 1 = 50

Successive division by 3 will give us the highest power of 3 that divides 55!:

55/3 = 18, 18/3 = 6, 6/3 = 2.

Highest power of 3 that divides 55! = 18 + 6 + 2 = 26.

So, 55! is a multiple of 2^50 and 3^26. What is the highest power of 12 that will divide

55!. Or, what is the highest value n can take such that (2^2 * 3)^n is a factor of 55!

We haev 26 threes's; but we can accommodate 2^2 only 25 times. Or, 12^25 will be a

factor of 55!, while 12^26 will not, since we need 52 2's for 12^26. We have only 50 2's.

Thus 12^25 will divide 55! - there are 25 trailing zeros in base 12 representation of 55!.

Question

A bus service runs from Chennai to Blr every third day with the first bus starting on Jan

1st. Another bus service runs from Chennai to Mumbai every 5th day starting from Jan

2nd. A third bus service from Chennai to Cochin runs on every 7th day starting from Jan

4th. In that year (which is not a leap year), on how many different days will a bus run

from Chennai to all three cities?

Explanation

Let us call Jan 1st as day 1, jan 2nd as day 2 and so on.

So,

The Chennai-Blr services runs on days 1, 4, 7, 10, 13, 16, 19,......

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The Chennai-Mumbai services runs on days 2, 7, 12, 17, 22, 27,......

The Chennai-Cochin services runs on days 4, 11, 18, 25, ......

First up, let us see if we can find one day where all three buses ply.

Chennai Blr runs on days of the form 3a + 1, Chennai_Mumbai on 5b + 2, and Chennai-

Cochin on 7c + 4. If N can be written as 3a + 1, 5b + 2 and 7c + 4, then we should be able

to get n as 105d + ___ . (105 is the LCM of 3, 5 and 7)

This is a very important idea. Get lots of practice on this idea. Wrap your head around

this idea very clearly.

First let us combine 3a + 1 and 5b + 2.

A number of the form 3a + 1, on division by 15 can have one of the following remainders

{ 1, 4, 7, 10, 13}

A number of the form 5b + 2, on division by 15 can have one of the following remainders

{ 2, 7, 12}

Or, if a number is of the form 3a + 1 and 5b + 2, it has to be of the form 15k + 7.

Now, N = 15K + 7 and 7c + 4.

Number 15k + 7, on division by 105 can have one of the following remainders { 7, 22,

37, 52, 67, 82, 97}}

Number 7c + 4, on division by 105 can have one of the following remainders { 4, 11, 18,

25, 32, 39, 46, 53, 60, 67, 74, 81, 88, 95, 102, 109}

Or, the number is of the form 105d + 67.

So, all three services will run on days 67, 67 + 105, 67 + 105*2 and so on.

Or, on days 67, 172 and 277. (Any other day would go beyond 365)

So, all three services would run together 3 days of the year.

Wonderful question testing concepts of LCM and Remainders.

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Question

How many positive integer values can x take that satisfy the inequality (x - 8) (x - 10) (x -

12).......(x - 100) < 0?

Answer: 30

Explanation

Let us try out a few values to see if that gives us anything.

When x = 8, 10, 12, ....100 this goes to zero. So, these cannot be counted.

When x = 101, 102 or beyond, all the terms are positive, so the product will be positive.

So, straight-away we are down to numbers 1, 2, 3, ...7 and then odd numbers from there

to 99.

Let us substitute x =1,

All the individual terms are negative. There are totally 47 terms in this list (How? Figure

that out). Product of 47 negative terms will be negative. So, x = 1 works. So, will x =2, 3,

4, 5, 6, and 7.

Remember, product of an odd number of negative terms is negative; product of even

number of negative terms is positive. Now, this idea sets up the rest of the question.

When x = 9, there is one positive terms and 46 negative terms. So, the product will be

positive.

When x = 11, there are two positive terms and 45 negative terms. So, the product will be

negative.

When x = 13, there are three positive terms and 44 negative terms. So, the product will be

positive.

and so on.

Essentially, alternate odd numbers need to be counted, starting from 11.

So, the numbers that will work for this inequality are 1, 2, 3, 4, 5, 6, 7...and then 11, 15,

17

19, 23, 27, 31,..... and so.

What will be the last term on this list?

99, because when x = 99, there are 46 positive terms and 1 negative term.

So, we need to figure out how many terms are there in the list 11, 15, 19,....99. These can

be written as

4 * 2 + 3,

4 * 3 + 3,

4 * 4 + 3

4 * 5 + 3

4 * 6 + 3

...

4 * 24 + 3

A set of 23 terms. So, total number of values = 23 + 7 = 30. 30 positive integer values of

x exist satisfying the condition.

Question: A merchant can buy goods at the rate of Rs. 20 per good. The particular good is part of an

overall collection and the value is linked to the number of items that are already on the

market. So, the merchant sells the first good for Rs. 2, second one for Rs. 4, third for Rs.

6…and so on. If he wants to make an overall profit of at least 40%, what is the minimum

number of goods he should sell?

A. 24

B. 18

C. 27

D. 32

Correct Answer: (C)

Explanation: Let us assume he buys n goods.

Total CP = 20n

Total SP = 2 + 4 + 6 + 8 ….n terms

Total SP should be at least 40% more than total CP

2 + 4 + 6 + 8 ….n terms > 1.4 * 20 n

2 (1 + 2 + 3 + ….n terms) > 28n

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n(n + 1) > 28n

n2 + n > 28n

n2 - 27n > 0

n > 27

He should sell a minimum of 27 goods.

Answer Choice (C)

Question:

P is x% more than Q. Q is (x - 10)% less than R. If P > R, what is the range of values x

can take?

A. 10% to 28%

B. 10% to 25%

C. 10% to 37%

D. 10% to 43%

Correct Answer: (C)

Explanation:

P = Q

Q = R

R =

P > R

(100 + x) (110 – x) > 100 x 100

11,000 + 110x – 100x – x2 > 10000

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1000 + 10x – x2 > 0

x2 – 10x – 1000 < 0

x2 – 10x + 25 < 1000 + 25

(x – 5)2 < 1025

x – 5 < 32

x < 37

x could range from 10% to 37%

Answer Choice (C)

Question: How many numbers with distinct digits are possible product of whose digits is 28?

A. 6

B. 4

C. 8

D. 12

Correct Answer: (C)

Explanation:

Two digit numbers; The two digits can be 4 and 7: Two possibilities 47 and 74

Three-digit numbers: The three digits can be 1, 4 and 7: 3! Or 6 possibilities.

We cannot have three digits as (2, 2, 7) as the digits have to be distinct.

We cannot have numbers with 4 digits or more without repeating the digits.

So, there are totally 8 numbers.

Answer Choice (C)

20

Question: If log2X + log4X = log0.25and x > 0, then x is

A. 6-1/6

B. 61/6

C. 3-1/3

D. 61/3

Correct Answer: (A)

Explanation:

log2x + log4x = log0.25

log2x + = log0.25

log2x * = log0.25

log2x * 3 = 2log0.25

log2x3 = log0.256

log2x3 = -log46

log2x3 =

log2x3 =

2log2x3 = -log26

2log2x3 + log26 = 0

log26x6 = 0

6x6 = 1

x6 =

x = Choice (A).

21

Question: 3x + 4|y| = 33. How many integer values of (x, y) are possible?

A. 6

B. 3

C. 4

D. More than 6

Correct Answer: (D)

Solution: Let us rearrange the equation:

3x = 33 – 4|y|

Since x and y are integers, and since |y| is always positive regardless of the sign of y, this

means that when you subtract a multiple of 4 from 33, you will get a multiple of 3.

Since 33 is already a multiple of 3, in order to obtain another multiple of 3, you will have

to subtract a multiple of 3 from it. So, y has to be a positive or a negative multiple of 3.

y = 3, -3, 6, -6, 9, -9, 12, -12...etc.

For every value of y, x will have a corresponding integer value.

So there are infinite integer values possible for x and y.

Question: 3sinx + 4cosx + r is always greater than or equal to 0. What is the smallest value ‘r’ can

to take?

A. 5

B. -5

C. 4

D. 3

Correct Answer: (A)

Explanation: 3sinx + 4cosx ≥ -r

≥ -r

= cosA => sinA =

5(sinx cosA + sinA cosA) ≥ -r

5(sin(x + A) ≥ -r

22

5sin (x + A) ≥ -r

-1 < sin (angle) < 1

5sin (x + A) ≥ -5

rmin = 5

Question: Solve the inequality x3 – 5x2 + 8x – 4 > 0.

A. (2, )

B. (1, 2) U (2, )

C. (-, 1) U (2, )

D. (-, 1)

Correct Answer : (B)

Explanation:

Let a, b, c be the roots of this cubic equation

a + b + c = 5

ab + bc + ca = 8

abc = 4

This happens when a = 1, b = 2 and c = 2 {This is another approach to solving cubic

equations}

The other approach is to use polynomial remainder theorem

If you notice, sum of the coefficients = 0

=> P(1) = 0

=> (x - 1) is a factor of the equation. Once we find one factor, we can find the other two

by dividing the polynomial by (x - 1) and then factorizing the resulting quadratic

equation.

(x - 1) (x - 2) (x - 2) > 0

Let us call the product (x - 1)(x - 2)(x - 2) as a black box.

If x is less than 1, the black box is a –ve number

If x is between 1 and 2, the black box is a +ve number

If x is greater than 2, the black box is a +ve number

23

Since we are searching for the regions where black box is a +ve number, the solution is as

follows:

1 < x < 2 OR x > 2

Answer Choice (B)

Question: An alloy of copper and aluminum has 40% copper. An alloy of Copper and Zinc has

Copper and Zinc in the ratio 2: 7. These two alloys are mixed in such a way that in the

overall alloy, there is more aluminum than Zinc, and copper constitutes x% of this alloy.

What is the range of values x can take?

A. 30% < x < 40%

B. 32.5% < x < 42%

C. 33.33% < x < 40%

D. 32.5% < x < 40%

Correct Answer : (A)

Explanation:

Alloy 1 Copper Aluminium

2x 3x

Alloy 2 Copper Zinc

2y 7y

If Aluminium and Zinc have to be equal, 3x = 7y. The simplest case occurs at the LCM,

when both Aluminium and Zinc are 21kg.

Alloy 1 Copper Aluminium

14 21

Alloy 2 Copper Zinc

6 21

Even a gram of alloy 1 above this level would mean that there is more Aluminium than

Zinc in the total alloy. So, this table is a limit on the percentages of different metals.

The percentage of Copper in the total alloy would be (14 + 6)* = 32.25%

24

On the other hand, if we want more Aluminium than Zinc, we can use tons and tons of

alloy 1, and only a microgram of Alloy 2. The quantity of alloy 2 can be made so small

that its presence can be almost neglected, as it will have as good as zero impact on the

overall percentage of copper. This is another extreme case, in which Copper will have

40% weight in the alloy.

Thus, the percentage of Copper ranges from 32.25% to 40% in this alloy.

Answer Choice (A)

Probability Question

Question: [x] = greatest integer less than equal to x. If x lies between 3 and 5. What is the

probability than [x2] = [x]2?

A. Roughly 0.64

B. Roughly 0.5

C. Roughly 0.14

D. Roughly 0.36

Correct Answer: (C)

Explanation:

Let us take a few examples.

[32] = [3]2

[3.52] =12 [3.5]2 = 9

[42] = 16 [4]2 = 16

For x (3, 5). [x]2 can only take value 9, 16 and 25.

Let us see when [x2] will be 9, 16 or 25

If [x2] = 9, x2 [9, 10)

=> x[3, )

[x2] = 16 => x2 [16, 17)

=> x [4, )

In the given range [x2] = 25 only when x = 5

So [x2] = [x]2 when x [3, ] or [4, ) or 5.

Probability =

= = 0.14

Questions and solutions on Probability

25

Question: John looks at the clock at some arbitrary time between 4pm and 6 pm and notices that the

angle between the two hands is an acute angle. He looks at the clock exactly 15 minutes

later, what is the probability that during this observation also he sees that the angle

between the hour and minute hands to be an acute-angle?

A. 1/2

B. 6/11

C. 13/24

D. 11/19

Correct Answer : (C)

Explanation:

At 4:00 pm the angle between the hands is 1200. The minutes hand is 1200 behind the

hour hand. Now, let us first see when the angle will become an acute angle. The angle

between he hands will be acute when the minutes hand gains more than 300. Or, after

30/330 * 60 minutes.

Somewhere between 4:05 and 4: 10.From now, the angle will be acute till the minute

hand reaches a point where it is 90 degree ahead of the hour hand. Or, till the time the

minute and gains 1800 (Going from 900 degree behind to 900 ahead of hour hand), the

angle will be acute. Or, for a spell of 180/330 * 60 minutes, the angle will be acute. Or

for a spell of 360/11 minutes, the angle will be acute. For a spell of 32 8/11 minutes the

angle will be acute.

So, if John had seen the clock at a time from 4 pm to 5pm, he would have seen it during

this spell. Now, if his second viewing had also happened within this spell, he would have

seen an acute angle second time.

Or, t, t + 15 should have both been within this 32 8/11 range for him to have seen two

acute angle observations. Or, t should have happened at least 15 minutes before the angle

become obtuse again. If we have the time range for it being acute as (x, x + 32 8/11), The

question can be restated as if t belongs to this range, what it is the probability that t + 15

also belongs to this range.

For t + 15 to be within the range, t + 15 < x + 32 8/11, or t < 32 8/11 – 15 = 17 8/11

Probability should be 195/11 divided by 360/11 = 195/360 = 39/72 = 13/24

26

In the 5pm to 6pm range we get an identical probability. So the overall probability is

13/24

Question: John was born on Feb 29th of 2012 which happened to be a Wednesday. If he lives to be

101 years old, how many birthdays would he celebrate on a Wednesday?

A. 3

B. 4

C. 5

D. 1

Correct Answer : (B)

Explanatory Answer: Let us do this iteratively. Feb 29th 2012 = Wednesday => Feb 28th 2012 = Tuesday

Feb 28th 2013 = Thursday (because 2012 is a leap year, there will be 2 odd days)

Feb 28th 2014 = Friday, Feb 28th 2015 = Saturday, Feb 28th 2016 = Sunday, Feb 29th

2016 = Monday

Or, Feb 29th to Feb 29th after 4 years, we have 5 odd days

So, every subsequent birthday, would come after 5 odd days.

2016 birthday – 5 odd days

2020 birthday – 10 odd days = 3 odd days

2024 birthday – 8 odd days = 1 odd day

2028 – 6 odd days

2032 – 11 odd days = 4 odd days

2036 – 9 odd days = 2 odd days

2040 – 7 odd days = 0 odd days. So, after 28 years he would have a birthday on

Wednesday

The next birthday on Wednesday would be on 2068 (further 28 years later), the one after

that would be on 2096. His 84th birthday would again be a leap year.

Now, there is a twist again, as 2100 is not a leap year. So, he does not have a birthday in

2100. His next birthday in 2104 would be after 9 odd days since 2096, or 2 odd days

since 2096, or on a Thursday.

From now on the same pattern continues. 2108 would be 2 + 5 odd days later = 7 odd

days later. Or, 2108 Feb 29th would be a Wednesday.

So, there are 4 occurrences of birthday falling on Wednesday – 2040, 2068 and 2096,

2108.

27

Question What is the remainder when (13100 +1 7100) is divided by 25?

A. 2

B. 0

C. 15

D. 8

Correct Answer: Choice (A)

Explanation:

What is the remainder when (13100 +1 7100) is divided by 25?

(13100 +1 7100) = (15 – 2)100 + (15 + 2)100

Now 52 = 25, So, any term that has 52 or any higher power of 5 will be a multiple of 25.

So, for the above question, for computing remainder, we need to think about only the

terms with 150 or 151.

(15 – 2)100 + (15 + 2)100

Coefficient of 150 = (-2)100 + 2100

Coefficient of 151 = 100C1 * 151* (-2)99 + 100C1 * 151* (-2)99 . These two terms cancel each

other.So, the sum is 0.

Remainder is nothing but (-2)100 + 2100 = (2)100 + 2100

2101

Remainder of dividing 21 by 25 = 2




Remainder of dividing 25 by 25 = 32 = 7

Remainder of dividing 210 by 25 = 72 = 49 = -1

Remainder of dividing 220 by 25 = (-1)2 = 1

Remainder of dividing 2101 by 25 = Remainder of dividing 2100 by 25 * Remainder of

dividing 21 by 25 = 1 * 2 = 2

Question A drain pipe can drain a tank in 12 hours, and a fill pipe can fill the same tank in 6 hours.

A total of n pipes – which include a few fill pipes and the remaining drain pipes – can fill

the entire tank in 2 hours. How many of the following values could ‘n’ take?

a) 24

b) 16

28

c) 33

d) 13

e) 9

f) 8

A. 3

B. 4

C. 2

D. 1

Correct Answer: (A)

Explanatory Answer: Two drain pipes can drain the same volume that one fill pipe fills. This means that a D-D-

F combination has to have a net volume effect of 0.

In spite of this, the tank still gets filled. Only the fill pipes can manage to fill the tank. In

addition to all the net zero effect pipes, we need three more fill pipes in order to fill the

tank in 2 hours.

So, we can have as many D-D-Fs as we want, but we need one F-F-F at the end to ensure

that the tank gets filled in 2 hours.

So the number of pipes will be → (D – D - F).......(D – D - F) + (F – F - F)

The number of pipes has to be a multiple of 3. Only options A, C and E fit the

description.

Answer Choice (A)

Question

29

Traders A and B buy two goods for Rs. 1000 and Rs. 2000 respectively. Trader A marks

his goods up by x%, while trader B marks his goods up by 2x% and offers a discount of

x%. If both make the same profit, find x

A. 25%

B. 12.5%

C. 37.5%

D. 40%

Answer Choice: (A)

Explanatory Answer:

SP of trader A = 1000 (1 + x)

Profit of trader A = 1000 (1 + x) - 1000

MP of trader B = 2000 (1 + 2x)

SP of trader B = 2000 (1 + 2x) (1 - x)

Profit of trader B = 2000(1 + 2x) (1 - x)- 2000

Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 - x)- 2000

1000x = 2000 – 4000x2 + 4000x – 2000x - 2000

4000x2 -1000x = 0

1000x (4x - 1) = 0

x = 25%

Answer Choice (A)

Question

(|x| -3) (|y| + 4) = 12. How many pairs of integers (x, y) satisfy this equation?

A. 4

30

B. 10

C. 6

D. 8

Answer: Choice (B)

Explanatory Answer

Product of two integers is 12. Further |y| + 4 > 4 . So, we know that one of the numbers

has to be greater than 4.

So, we can have

1 x 12 => x = +4, y = +8

2 x 6 => x = +5, y = +2

3 x 4 => x = +6, y = 0

4 + 4 + 2 = 10 possible solutions

Question

How many of the following statements have to be true?

I. No year can have 5 Sundays in the month of May and 5 Thursdays in the

month of June

II. If Feb 14th of a certain year is a Friday, May 14th of the same year cannot be

a Thursday

III. If a year has 53 Sundays, it can have 5 Mondays in the month of May

A. 0

B. 1

C. 2

D. 3

31

Answer: Choice (B)

Explanation

I. No year can have 5 Sundays in the month of May and 5 Thursdays in the

month of June

A year has 5 Sundays in the month of May => it can have 5 each of Sundays, Mondays

and Tuesdays, or 5 each of Saturdays, Sundays and Mondays, or 5 each of Fridays,

Saturdays and Sundays. Or, the last day of the Month can be Sunday, Monday or

Tuesday.

Or, the 1st of June could be Monday, Tuesday or Wednesday. If the first of June were a

Wednesday, June would have 5 Wednesdays and 5 Thursdays. So, statement I need not

be true.

II. If Feb 14th of a certain year is a Friday, May 14th of the same year cannot

be a Thursday

From Feb 14 to Mar 14, there are 28 or 29 days, 0 or 1 odd day

Mar 14 to Apr 14, there are 31 days, or 3 odd datys

Apr 14 to May 14 there are 30 days or 2 odd days

So, Feb 14 to May 14, there are either 5 or 6 odd days

So, if Feb 14 is Friday, May 14 can be either Thursday or Wednesday. So, statement 2

need not be true.

III. If a year has 53 Sundays, it can have 5 Mondays in the month of May

Year has 53 Sundays => It is either a non-leap year that starts on Sunday, or leap year

that starts on Sunday or Saturday.

Non-leap year starting on Sunday: Jan 1st = Sunday, jan 29th = Sunday. Feb 5th is Sunday.

Mar 5th is Sunday, Mar 26 is Sunday. Apr 2nd is Sunday. Apr 30th is Sunday, May 1st is

Monday. May will have 5 Mondays.

So, statement C can be true.

Only one of the three statements needs to be true. Answer Choice (B)

Question

x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z.

A. 2

32

B. 3

C. 1

D. More than one possible value of z exists

Answer: Choice B

Explanatory Answer: xy = 4

xy could be 2 x 2 or 4 x 1

221

222 These are the possible triangles

223

441

22x will be a triangle if x is 1, 2 or 3 (trial and error)

44x is a triangle only if x is 1.

221 is acute. 12 + 22> 22

222 is equilateral. So acute.

223 is obtuse. 22 + 22< 32

144 is acute. 12 + 42> 42

Only triangle 223 is obtuse. Hence, the third side has to be 3.

Answer Choice (B)

Question

33

Sides of a triangle are 6, 10 and x for what value of x is the area of the the

maximum?

A. 8 cms

B. 9 cms

C. 12 cms

D. None of these

Correct Answer: D

Explanatory Answer

Side of a triangle are. 6, 10, x.

Area = 1/2 x 6 x 10 sin BAC.

Area is maximum, when BAC = 90o

x = sqrt(100 + 36) = sqrt(136)

There is a more algebraic method using hero’s formula. Try that also.

Answer Choice (D)

Question

Consider a class of 40 students whose average weight is 40 kgs. m new students join this

class whose average weight is n kgs. If it is known that m + n = 50, what is the maximum

possible average weight of the class now?

A. 40.18 kgs

B. 40.56 kgs

C. 40.67 kgs

34

D. 40.49 kgs

Correct Answer: Choice (B)

Explanation: If the overall average weight has to increase after the new people are added, the average

weight of the new entrants has to be higher than 40.

So, n > 40

Consequently, m has to be <10 (as n + m = 50)

Working with the “differences” approach, we know that the total additional weight added

by “m” students would be (n - 40) each, above the already existing average of 40. m(n -

40) is the total extra additional weight added, which is shared amongst 40+m students.

So, m (n – 40)/ (m + 40) has to be maximum for the overall average to be maximum.

At this point, use the trial and error approach (or else, go with the answer options) to

arrive at the answer.

The maximum average occurs when m = 5,and n = 45

And the average is 40 + (45 – 5) * 5/45 = 40 + 5/9 = 40.56

Answer Choice (B)

Question

1. 5. Scores in a classroom are broken into 5 different ranges, 51-60, 61-70, 71-80, 81-90

and 91-100. The number of students who have scored in each range is given below

51 -60 - 3 students

61 -70 - 8 students

71 -80 - 7 students

81 -90 - 4 students

91 -100 - 3 students

35

Furthermore, we know that at least as many students scored 76 or more as those who

scored below 75. What is the minimum possible average overall of this class?

A. 72

B. 71.2

C. 70.6

D. 69.2

Answer: Choice (C)

Explanatory Answer:

Let's employ the idea of a total of 25 students (all of the same weight) sitting on a see-

saw, which has numbers from 51 to 100 marked on it. At least as many students are

sitting on 76 (or to its right), as there are sitting to the left of 75. Now this means that you

can have only one person sitting to the left of 75 and all the rest sitting beyond 76. But

you can't do that, as you have other constraints as well.

First of all, you have to seat 3 students from 51 to 60, and 8 students from 61 to 70.

Secondly, you also have to make sure that the average is the least. This means that the

see-saw should be tilting as much to the left as possible, which in turn means that the

number of people sitting to the left of 75 should be the highest possible.

This makes it 12 students to the left of 75, and the remaining 13 students on 76 or to its

right.

Next, how do you ensure that the average is least, i.e. how do you ensure that the balance

tilts as much as possible to the left? Make each student score as little as possible given

the constraints.

36

So, the first 3 students only score 51 each. The next 8 students score only 61 each. 11

Students are now fixed. The 12th student has to be below 75, so seat him on 71. The

remaining 6 students (who are in the 71 to 80 range) have to score 76. The next 4 score

81 and the next 3 score 91.

This would give you the least average.

The lowest possible average would be:

= [(3 * 51) + (8 * 61) + 71 + (6 * 76) + (4* 81) + (3 * 91)]/25

= 70.6

Answer Choice (C)

Question

How many onto functions can be defined from the set A = {1, 2, 3, 4} to {a, b, c}?

A. 81

B. 79

C. 36

D. 45

Answer: Choice (C)

Explanatory Answer:

First let us think of the number of potential functions possible. Each element in A has

three options in the co-domain. So, the number of possible functions = 34 = 81.

37

Now, within these, let us think about functions that are not onto. These can be under two

scenarios

Scenario 1: Elements in A being mapped on to exactly two of the elements in B (There

will be one element in the co-domain without a pre-image).

Let us assume that elements are mapped into A and B. Number of ways in which this can

be done = 24 – 2 = 14

o 24 because the number of options for each element is 2. Each can be mapped on to either

A or B

o -2 because these 24 selections would include the possibility that all elements are mapped

on to A or all elements being mapped on to B. These two need to be deducted

The elements could be mapped on B & C only or C & A only. So, total number of

possible outcomes = 14 * 3 = 42.

Scenario 2: Elements in A being mapped to exactly one of the elements in B. (Two

elements in B without pre-image). There are three possible functions under this scenario.

All elements mapped to a, or all elements mapped to b or all elements mapped to c.

Total number of onto functions = Total number of functions – Number of functions

where one element from the co-doamin remains without a pre-image - Number of

functions where 2 elements from the co-doamin remain without a pre-image

Total number of onto functions = 81 – 42 – 3 = 81 – 45 = 36

Answer Choice (C)

38

Question

1. x + |y| = 8, |x| + y = 6.How many pairs of x, y satisfy these two equations?

A. 2

B. 4

C. 0

D. 1

Answer: Choice (D)

Explanatory Answer We start with the knowledge that the modulus of a number can never be negative, though

the number itself may be negative.

The first equation is a pair of lines defined by the equations

y = 8 – x ------- (i) {when y is positive}

y = x – 8 ------- (ii) {when y is negative}

With the condition that x ≤ 8 (because if x becomes more than 8, |y| will be forced to be

negative, which is not allowed)

The second equation is a pair of lines defined by the equations:

y = 6 – x ------- (iii) {when x is positive}

y = 6 + x ------- (iv) {when x is negative}

with the condition that y cannot be greater than 6, because if y > 6, |x| will have to be

negative.

On checking for the slopes, you will see that lines (i) and (iii) are parallel. Also (ii) and

(iv) are parallel (same slope).

39

Lines (i) and (iv) will intersect, but only for x = 1; which is not possible as equation (iv)

holds good only when x is negative

Lines (ii) and (iii) do intersect within the given constraints. We get x = 7, y = -1. This

satisfies both equations.

Only one solution is possible for this system of equations.

Question

Qn. f(x + y) = f(x)f(y) for all x, y, f(4) = +3 what is f(-8)?

A. 1/3

B. 1/9

C. 9

D. 6

Answer Choice: B

Explanatory Answer

f(x + 0) = f(x) f(0)

f(0) = 1

f(4 + -4) = f(0)

f(4 + -4) = f(4) f(-4)

1 = +3 x f(-4)

f(-4) = 1/3

f(-8) = f(-4 + (-4)) = f(-4) f(-4)

f(-8) = 1/3 x 1/3 = 1/9

Answer Choice (B)

40

Question

Second term of a GP is 1000 and the common ratio is where n is a natural number. Pn is

the product of n terms of this GP. P6 > P5 and P6 > P7, what is the sum of all possible

values of n?

A. 5

B. 9

C. 8

D. 11

Correct Answer

Choice B

Explanatory Answer

Common ratio is positive, and one of the terms is positive => All terms are positive

P6 = P5 * t6 => If P6 > P5, t6 > 1

P7 = P6 * t7 => If P6 > P7, t7 < 1

T6 = t2 * r4 = 1000r4;

T7 = t2 * r5 = 1000r5

1000r4 > 1 and 1000r5 < 1

1/r^4 < 1000 and 1/r^5 > 1000

1/r = n

n4 < 1000 and n5 > 1000, where n is a natural number

n4 < 1000 => n < 6

n5 > 1000 => n > 4

n could be 4 or 5. Sum of possible values = 9

Answer Choice (B)

Question

41

If all words with 2 distinct vowels and 3 distinct consonants were listed alphabetically,

what would be the rank of “ACDEF’?

A. 4716

B. 4720

C. 4718

D. 4717

Correct Answer

Choice C

Explanatory Answer

The first word would be ABCDE. With 2 distinct vowels, 3 distinct consonants, this is

the first word we can come up with.

Starting with AB, we can have a number of words

AB __ __ __. The next three slots should have 2 consonants and one vowel. This can be

selected in 20C2 and 4C1 ways. Then the three distinct letters can be rearranged in 3!

Ways.

Or, number of words starting with AB = 20C2 * 4C1 * 3! = 190 * 4 * 6 = 4560

Next, we move on to words starting with ACB

ACB __ __. The last two slots have to be filled with one vowel and one consonant. = 19C1

* 4C1. This can be rearranged in 2! Ways.

Or, number of words starting with ACB = 19C1 * 4C1 * 2 = 19 * 4 * 2 = 152

Next we move on words starting with ACDB: There are 4 different words on this list –

ACDBE, ACDBI, ACDBO, ACDBU

So, far number of words gone = 4560 + 152 + 4 = 4716

Starting with

AB 4560

Starting with

ACB 152

Starting with 4

42

ACDB

Total words

gone 4716

After this we move to words starting with ACDE, the first possible word is ACDEB.

After this we have ACDEF.

So, rank of ACDEF = 4718

Answer Choice (C)

Question

If we listed all numbers from 100 to 10,000, how many times would the digit 3 be

printed?

A. 3980

B. 3700

C. 3840

D. 3780

Correct Answer

Choice A

Explanatory Answer

We need to consider all three digit and all 4-digit numbers.

Three-digit numbers: A B C. 3 can be printed in the 100’s place or10’s place or units

place.

Ø 100’s place: 3 B C. B can take values 0 to 9, C can take values 0 to 9. So, 3 gets printed

in the 100’s place 100 times

Ø 10’s place: A 3 C. A can take values 1 to 9, C can take values 0 to 9. So, 3 gets printed in

the 10’s place 90 times

Ø Unit’s place: A B 3. A can take values 1 to 9, B can take values 0 to 9. So, 3 gets printed

in the unit’s place 90 times

43

So, 3 gets printed 280 times in 3-digit numbers

Four-digit numbers: A B C D. 3 can be printed in the 1000’s place, 100’s place or10’s

place or units place.

Ø 1000’s place: 3 B C D. B can take values 0 to 9, C can take values 0 to 9, D can take

values 0 to 9. So, 3 gets printed in the 100’s place 1000 times.

Ø 100’s place: A 3 C D. A can take values 1 to 9, C & D can take values 0 to 9. So, 3 gets

printed in the 100’s place 900 times.

Ø 10’s place: A B 3 D. A can take values 1 to 9, B & D can take values 0 to 9. So, 3 gets

printed in the 10’s place 900 times.

Ø Unit’s place: A B C 3. A can take values 1 to 9, B & C can take values 0 to 9. So, 3 gets

printed in the unit’s place 900 times.

3 gets printed 3700 times in 4-digit numbers.

So, there are totally 3700 + 280 = 3980 numbers

Answer Choice (A)

Question

A is x% more than B and is x% of sum of A and B. What is the value of x? Give

approximate answer.

Correct Answer

Roughly 62%

Explanatory Answer

a = b (1 + x) => a/b = 1 + x

a = x (a + b) , dividing by a through out

1 = x (1 + b/a)

1 = x (1 + 1/(1 + x))

1 = x (x + 2) / (x + 1)

x + 1 = x2 + 2x

44

=> x2 + x - 1 = 0

Now, we need to solve this equation. Using the discriminant method, when we solve this,

x turns out to be {-1 + sqrt (5)}/2

x has to lie between 0 and 1 and therefor cannot be { -1 - sqrt (5) }/2.

So, the only solution is { -1 + sqrt (5) }/2. This is roughly 0.62.

Or, x has to be 62% approximately. The ration 1.618 is also called the golden ratio, and is

the conjugate and reciprocal of 0.618.

The golden ratio finds many mentions, from the Fibonacci series to Da Vinci. So, it is a

big favourite of mathematician.

Question

The sum of the factors of a number is 124. What is the number?

Question

The sum of the factors of a number is 124. What is the number?

Correct Answer

Number could be 48 or 75

Explanatory Answer

This video gives the solution for this question. Given below the video is the explanation

(in words)

Any number of the form paqbrc will have (a+1) (b+1)(c+1) factors, where p, q, r are

prime. (This is a very important idea)

For any number N of the form paqbrc, the sum of the factors will be (1 + p1 + p2 + p3+ …+

pa) (1 + q1 + q2 + q3+ …+ qb) (1 + r1 + r2 + r3+ …+ rc).

Sum of factors of number N is 124. 124 can be factorized as 22 * 31. It can be written as

4 * 31, or 2 * 62 or 1 * 124.

2 cannot be written as (1 + p1 + p2 …pa) for any value of p.

4 can be written as (1 + 3)

So, we need to see if 31 can be written in that form.

The interesting bit here is that 31 can be written in two different ways

31 = (1 + 21 + 22+ 23 + 24)

http://en.wikipedia.org/wiki/Golden_ratio

45

31 = ( 1 + 5 + 52)

Or, the number N can be 3 * 24 or 3 * 52. Or N can be 48 or 75.

Question

Consider functions f(x) = x2 + 2x, g(x) = √(x +1) and h(x) = g(f(x)). What are the

domain and range of h(x)?

Correct Answers:

Domain = ( - infinity, +infinity)

Range - [0, infinity]

Explanatory Answer

h(x) = g(f(x)) = g(x2 + 2x) = ( x2 + 2x + 1) = (x+1)2 = | x + 1|

This bit is very important, and often overlooked

sqrt(9) = 3, not +3

If x2 = 9, then x can be +3, but 9) is only +3.

So, x2)= |x|, not +x, not + x

Domain of | x + 1| = ( -infinity, + infinity), x can take any value.

As far as the range is concerned, | x + 1| cannot be negative. So, range = [0, infinity)

Question

log5x = a (This should be read as log X to the base 5 equals a)log20x = b

What is logx10 ?

Correct Answer: (a + b)/2ab

Explanation

log5x = a

log20x = b

logx5 = 1/a

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logx20 = 1/b

logx100/5 = 1/b

logx100 – logn5 = 1/b

2logx10 – 1/a = 1/b

logx10 = ½(1/a + 1/b)

logx10 = (a + b)/2ab

Questions

1. Find the range of x for which (x + 2) (x + 5) > 40.

2. How many integer values of x satisfy the equation x( x + 2) (x + 4) (x + 6) < 200

3. Find the range of x where ||x - 3| - 4| > 3.

Correct Answers

1. x < -10 or x > 3

2. 9 different values

3. (-infinity, -4) or (2,4) or ( 10, infinity)

Explanatory Answers

1. Find the range of x for which (x + 2) (x + 5) > 40.

There are two ways of trying this one. We can expand and simplify this algebraically. x^2

+ 7x + 10 > 40 or

x^2 + 7x - 30 > 0

(x + 10) (x -3) > 0

The roots are -10 and +3.

=> x should lie outside the roots.

Now, what is this based on?

There is a simple thumb rule for solving quadratic inequality

For any quadratic inequality ax2 + bx + c < 0

Factorize it as a(x - p) ( x - q) < 0

Whenever a is greater than 0, the above inequality will hold good if x lies between p and

q.

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a(x - p) (x - q) will be greater than 0, whenever x does not lie between p and q. In other

words x should lie in the range ( -infinity, p) or (q, infinity)

Now, coming back to the question (x+10) (x -3) > 0

Or,x < -10 or x > 3

Second method: 5 * 8 = 40, -8 * -5 = 40

So, if x + 2 > 5 this will hold good => x > 3

If x + 2 is less than -8 also, this will hold good => x < -10. The first method is far more

robust.

2. How many integer values of x satisfy the equation x( x + 2) (x + 4) (x + 6) < 200

To begin with - 0, -2, -4 and -6 work. These are the values for which the left-hand side

goes to zero.

There are 4 terms in the product. If all 4 are positive or all 4 are negative the product will

be positive

The product can be negative only if exactly 1 or exactly 3 are negative. When 1 or 3

terms are negative, the product is clearly less than 200.

When x = -1, one term is negative

When x = -5, three terms are negative

So, adding these two numbers also to the set of solutions {-6, -5, -4, -2, -1, 0} satisfy the

inequality.

Beyond this it is just trial and error.

Let us try x = 1. Product is 1 * 3 * 5 * 7 = 105. This works

x = -7 gives the same product. So, that also works.

So, the solution set is now refined to {-7, -6, -5, -4, -2, -1, 0, 1}

x = 2 => Product is 2 * 4 * 6 * 8 = 8 * 48. Not possible. Any x greater than 1 does not

work

x = -8 is also not possible. Any value of x less than -7 does not work.

48

So, the solution set stays as {-7, -6, -5, -4, -2, -1, 0, 1}

The one missing value in this sequence is -3. When x = -3, product becomes -3 * -1 * 1 *

3. = 9. This also holds good.

So, values {-7, -6,-5, -4, -3, -2, -1, 0, 1} hold good. 9 different values satisfy this

inequality

3. Find the range of x where ||x - 3| - 4| > 3

If we have an inequality |y| > 3, this will be satisfied if => y > 3 or y < -3.

So, the above inequality simplifies to two inequalities

Inequality I: | x - 3| - 4 > 3 | x - 3| - 4 > 3 => | x - 3 | > 7

x - 3 > 7 or x - 3 < -7

Or, x > 10 or x < -4

Of, x lies outside of -4 and 10. Or, x can lie in the range ( - infinity, -4) or ( 10, infinity)

Inequality II: |x -3| - 4 < -3

|x - 3 | - 4 < - 3

=> | x - 3 | < 1

=> -1 < x - 3 < 1 or x lies between 2 and 4

So, final solution is given by the range

( - infinity, -4) or (2,4) or ( 10, infinity)

NOTE:

Competitive exams often ask questions with a 'wrapper' around them. Its important to get

to the right question quickly. Give the underlying questions for the following -

1. Give the smallest 4-digit number with an odd number of factors (easy one)

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2. Set S contains elements { 6300, 2100, 1260, 900, 700, 6300/11, 6300/13, 420, ..........},

how many elements of Set S are integers?

3. f(k) gives the sum of all digits of number k. g(k) = f(f(f(...k))) such that we end up with

a number from 1 to 9. How many 5-digit numbers n exist such that g(n) = 2

Discussion:

The questions sitting underneath the above statements are as follows

1. Give the smallest 4-digit number with an odd number of factors (easy one) => What is

the smallest 4-digit perfect square?


how many elements of Set S are integers? => How many odd factors does 6300 have?


a number from 1 to 9. How many 5-digit numbers n exist such that g(n) = 2? => How

many 5-digit number exist that when divided by 9 leave a remainder of 2?

Explanatory Answers

1. Give the smallest 4-digit number with an odd number of factors (easy one) => What is

the smallest 4-digit perfect square? 1024


how many elements of Set S are integers? => How many odd factors does 6300 have?

6300 = 2^2 * 3^2 * 5^2 * 7. Number of odd factors of this number = (2+1) * (2+1) *

(1+1) = 18. For discussion on number of odd factors, look here .


a number from 1 to 9. How many 5-digit numbers n exist such that g(n) = 2? => How

many 5-digit number exist that when divided by 9 leave a remainder of 2?

There are 90000 5-digit numbers. There will be 10000 numbers that leave a remainder of

2 on division by 9 within these. Answer = 10000.

http://iimcat.blogspot.in/2010/11/solutions-to-number-theory-questions.html

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g(n) mentioned above is identical to the remainder when a number is divided by 9. Once

you pick that, this question become a sitter. Competitive exams are good at masking

questions. As much as possible, learn from first principles. If one had thought about why

the test of divisibility for 9 works, this bit would have been clear.

NOTE 2;

Key formulae

For a1 + a2 + a3 ....ar = n. The number of solutions where a1,a2,a3....ar all take natural

numbers is (n-1) C (r-1).

For a1 + a2 + a3 ....ar = n. The number of solutions where a1,a2,a3....ar can take all

whole number values is (n+r-1) C (r-1).

Similar questions on the same theme

The proofs are just an extension of the idea discussed in the two posts. Now, this kind of

question can get asked in different frameworks. Few examples are given below.

1. 10 identical toys need to be placed in 3 distinct boxes. In how many ways can this be

done?

2. Mark needs to pick up exactly 40 fruits for his family. He has to pick at least 8 Apples,

at least 6 Oranges and at least 5 mangoes. He should not pick up any other fruit. In how

many ways can this be done?

Question

Sum of three whole numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?

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Correct Answer

12C2

Explanatory Answer

a + b + c = 10. a,b,c are whole numbers. Now this is similar to the previous question that

we solved by placing 10 sticks and simplifying. The discussion can be seen here .

We cannot follow a similar approach as a,b,c can be zero. Let us modify the approach a

little bit. Let us see if we can remove the constraint that a,b,c can be zero.

If we give a minimum of 1 to a,b,c then the original approach can be used. And then we

can finally remove 1 from each of a,b, c. So, let us distribute 13 sticks across a, b and c

and finally remove one from each.

a + b + c = 13. Now, let us place ten sticks in a row

| | | | | | | | | | | | |

This question now becomes the equivalent of placing two '+' symbols somewhere

between these sticks. For instance

| | | | + | | | | | + | | | |, This would be the equivalent of 4 + 5 + 4.

or, a = 4, b = 5, c = 4.

There are 12 slots between the sticks, out of which one has to select 2 for placing the '+'s.

The number of ways of doing this is 12C2.

http://iimcat.blogspot.in/2012/10/questions-on-permutation-and-combination.html

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Questions

1. Sum of three natural numbers a, b and c is 10. How many ordered triplets (a, b, c)

exist?

2. In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is

empty?

Correct Answers:

1. 9C2

2. 10C2

Explanatory Answers

1. Sum of three natural numbers a, b and c is 10. How many ordered triplets (a, b, c)

exist?

a + b + c = 10. Now, let us place ten sticks in a row

| | | | | | | | | |

This question now becomes the equivalent of placing two '+' symbols somewhere

between these sticks. For instance

| | | | + | | | | | + |, This would be the equivalent of 4 + 5 + 1. or, a = 4, b =

5, c = 1.

There are 9 slots between the sticks, out of which one has to select 2 for placing the '+'s.

The number of ways of doing this would be 9C2. Bear in mind that this kind of

calculation counts ordered triplets. (4,5,1) and (1, 4, 5) will both be counted as distinct

possibilities.

We can also do a + b + c = n where a, b, c have to be whole numbers (instead of natural

53

numbers as in this question) with a small change to the above approach. Give it some

thought.

2. In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is

empty?

This is nothing but number of ways of having a,b,c such that a + b + c = 11, where a, b, c

are natural numbers.

10C2.

1. (|x| - 2) ( x + 5) < 0. What is the range of values x can take?

2. a and b are roots of the equation x^2 - px + 12 = 0. If the difference between the roots

is at least 12, what is the range of values p can take?

Solutions to the above questions

1. (|x| - 2) ( x + 5) < 0 -

This can be true in two scenarios

Scenario I - (|x| - 2) < 0 and ( x + 5) > 0

Or |x| < 2 and x > -5.This gives us the range (-2,2)

Scenario II - (|x| - 2) > 0 and ( x + 5) < 0

Or |x| > 2 and x < -5. This gives us the range (-Infinity, -2)

So, the overall range is (-infinity, -2) or (-2,2)

2. a and b are roots of the equation x^2 - px + 12 = 0. If the difference between the

roots is at least 12, what is the range of values p can take?

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The roots are a and b

a + b = p ab = 12

(a + b )^2 = p^2

(a -b)^2 = (a + b ) ^2 - 4ab

=> (a-b) ^2 = p^2 - 12*4 = p^2 - 48

If |a-b| > 12 { Difference between the roots is at least 12}

then, (a-b)^2 > 144

p^2 - 48 > 144

p^2 > 192

P > 8sqrt(3) or P < -8 sqrt(3)

Questions:

1. Sum of three distinct natural numbers is 25. What is the maximum value of their

product?

2. x ( x + 3) ( x + 5) (x + 8) < 250. How many integer values can x take?

Correct Answers:

Question 1: 560

Question 2: 11 values

Explanatory Answers

Qn 1. Sum of three distinct natural numbers is 25. What is the maximum value of their

product?

Let the three natural numbers be a, b and c. We know that a + b + c = 25.

55

Let us substitute some values and see where this is headed

a = 1, b = 2 c = 22, Product = 44

a = 2, b = 3 c = 20, product = 120

a = 5, b = 6 c = 14, product = 420

The numbers should be as close to each other as possible. (Just trial and error and one can

figure this out). This property is an extension of the AM-GM Inequality.

25/3 ~ 8. 25 divided by 3 is approximately equal to 8. So, we should choose a, b, and c

close enough to 8

8 + 8 + 9 = 25. But the numbers need to be distinct. 8 * 7 * 10 come closest. The

maximum product would be 560.

Qn 2. x ( x + 3) ( x + 5) (x + 8) < 250. How many integer values can x take?

Straight away we can see that x = 0 works. The product goes to zero in this case. When x

takes values -3, -5 or -8, the product will go to zero and the inequality holds good.

Now, let us substitute some other values

When x = 1 => Product = 1 * 4 * 6 * 9 = 216 < 250 So, x =1 holds good

When x = 2, the product is clearly greater than 250.

So, thus far, we have seen that for x = 0 , -3 -5 -8 or 1; the above inequality holds good.

There are 4 terms in this product. If all 4 are positive or all 4 are negative, the product

will be positive. If exactly one term is positive or exactly one term is negative, the

product will be negative.

Whenever the product is negative, the inequality will hold good. So, let us find the values

of x for which the product will be negative

x = -1 or -2, the product is negative, so the inequality will hold good

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Let us think of other values of x for which the product is negative. For the product to be

negative, either 1 or 3 of the four terms should be negative.

When x is -6 or -7, three of the terms are negative and the product is negative.

So, for x = 0, 1, -3, -5, -8, -1, -2, -6 or -7 this holds good. We have seen 9 values thus far.

In ascending order, the values are -8, -7, -6, -5, -3, -2, -1, 0, 1

The value in between that we have thus far not verified is -4. Let us try -4 as well. In this

case the product is -4 * -1 * 1 * 7 < 250

For x = -9, product is -9* -6 * -4 * 1 = 216 < 250

Clearly for x = -10 or lesser the inequality does not hold good.

So, the inequality holds good for 11 terms. x can take all integers from -9 to +1 (both

inclusive).

1. Give the domain and range of the following functions

i. f(x) = x2 + 1

Domain = All real numbers (x can take any value)

Range [1, infinity). Minimum value of x2 is 0.

ii. g(x) = log (x + 1)

Domain = Log of a negative number is not defined so (x + 1) > 0 or x > -1

Domain ( -1, infinity)

Range = (-infinity, +infinity)

Note: Log is one of those beautiful functions that is defined from a restricted domain to

all real numbers. Log 0 is also not defined. Log is defined only for positive numbers

iii. h(x) = 2xDomain - All real numbers.

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Range = (0, infinity)

The exponent function is the mirror image of the log function.

iv. f(x) = 1/ (x+1)

Domain = All real numbers except -1

Range = All real numbers except 0

v. p(x) = | x + 1|

Domain = All real numbers

Range = [0, infinity) Modulus cannot be negative

vi. q(x) = [2x], where [x] gives the greatest integer less than or equal to x

Domain = All real numbers

Range = All integers

The range is NOT the set of even numbers. [2x] can be odd. [2*0.6] = 1. It is very

important to think fractions when you are substituting values.

1. Give the domain and range of the following functions

i. f(x) = x2 + 1

ii. g(x) = log (x + 1)

iii. h(x) = 2x

iv. f(x) = 1/ (x+1)

v. p(x) = | x + 1|

vi. q(x) = [2x], where [x] gives the greatest integer less than or equal to x

1. A point P is identified as P(m,n). What is the ratio AP:BP given that the points A

and B are identified as A(5,-4) and B(1,6)?

Stmt 1. m = 3. Not sufficient.

Stmt 2. n = 2.5 Not sufficient.

Either statement alone is not sufficient. Both put together, we can answer the

questions. Ans B

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2. A point P is identified as P(m,n). What is the ratio AP:BP given that the points

A and B are identified as A(5,-4) and B(5,6)?

Stmt 1. m = 5. Not sufficient. It tells us that all points lie on the line x = 5, but this

is not enough

Stmt 2. n = 3 Not sufficient.

Both put together, this is enough.

In the above question, would n = 1 have been sufficient? Think about that.

3. A survey of 100 people tried to find the number of people who can write with

both their left and right hands. What is the maximum number of people who could

write left-handed and right-handed?

Stmt 1. 50 people can write only with their left hand. 40 people can write only with

their right hand. Sufficient: Maximum of 10 people could write left-handed and

right-handed.

Stmt 2. 50 people can write with their left hand. 40 people can write with their

right hand. Sufficient: Maximum of 40 people could write left-handed and right-

handed.

Answer Choice C

4. What is the slope of a line?

Stmt 1: The line makes 135 degrees with the negative direction of x - axis.

Sufficient: The line makes 45 degrees with positive x axis. This should be enough.

Stmt 2: The line makes an isosceles right triangle with the coodinate axes and the

product of the intercepts is negative. Sufficient: Either both intercepts are positive

and equal or negative and equal. Slope = -1

1. Perimeter of a triangle with integer sides is equal to 15. How many such triangles

are possible?

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This is just a counting question, with the caveat that sum of two sides should be

greater than the third. Let us assume a < b < c

a = 1, Possible triangle 1, 7, 7

a = 2, possible triangle 2, 6, 7

a = 3, possible triangles 3, 6, 6 and 3, 5, 7

a = 4, possible triangles 4, 4, 7 and 4, 5, 6

Again, from comments,

a = 5, possible triangle is 5,5,5,

There are totally 7 triangles possible

2. Triangle ABC has integer sides x, y, z such that xz = 12. How many such

triangles are possible?

xz = 12

x,z can be 1, 12 or 2, 6 or 3, 4

Possible triangles

1-12-12

2-6-5; 2-6-6; 2-6-7

3-4-2; 3-4-3; 3-4-5; 3-4-6.

As pointed out in the comments section, I have missed the triangle 3-4-4.

There are totally 9 triangles.

3. Triangle has sides a^2, b^2 and c^2. Then the triangle with sides a, b, c has to be

- a) Right angled b) Acute-angled c) Obtuse angled d) can be any of these three

Assuming a < b < c, we have a^2 + b^2 > c^2. This implies the triangle with sides

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a, b, c has to be acute-angled.

P.S: Big thanks to 'maniac' for pointing out the errors





Exactly one of ab, bc and ca is odd => Two are odd and one is even

abc is a multiple of 4 => the even number is a multiple of 4

The arithmetic mean of a and b is an integer => a and b are odd

and so is the arithmetic mean of a, b and c. => a+ b + c is a multiple of 3

c can be 4 or 8.

c = 4; a, b can be 3, 5 or 5, 9

c = 8; a, b can be 3, 7 or 7, 9

Four triplets are possible


multiples of 12?

Number should be a multiple of 3 and 4. So, the sum of the digits should be a

multiple of 3. WE can either have all seven digits as 3, or have three 2's and four

3's, or six 2's and a 3. (The number of 2's should be a multiple of 3).

For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us

combine these two.

All seven 3's - No possibility

Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some

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order. No of possibilities = 5!/3!*2! = 10

Six 2's and one 3 - The first 5 digits should all be 2's. So, there is only one number

2222232



be a perfect square a,b,c have to be even. a can take values 0, 2, 4. b can take

values 0,2, 4, 6 and c can take values 0,2. Total number of perfect squares = 3 * 4 *

2 = 24



be an odd number, a should be 0. b can take values 0,1, 2,3, and c can take values

0, 1, 2,3, 4. Total number of odd factors = 4 * 5 = 20



Number can be 7, 25, 43, 61, 79.

Remainders when divided by 12 are 7 and 1.

NOTE 3






multiples of 12?

62





NOTE 4

1. Any two triangles that are congruent to each other will also be similar to each other

2. If in a triangle with sides a, b and c, if a^2 + b^2 > c^2 the triangle has to be

acute-angled

3. Any parallelogram inscribed inside a circle has to be a rectangle

4. If in a triangle the orthocenter, incenter and circumcenter are collinear the

triangle has to be isosceles.

5. There will be a unique circle passing through any three points

6. If two circles with centers A and B and radii r and R intersect, then AB > R - r

7. Circumradius of a triangle cannot be greater than the three sides of the triangle

8. For any obtuse-angled triangle, the orthocenter and circumcenter will lie outside

the triangle

9. In a scalene triangle, the sum of the three medians will be greater than the sum

of the three altitudes.

10. In circumcircle and incircle are concentric, the triangle has to be equilateral

1. From the digits 2,3,4,5,6 and 7, how many 5-digit numbers can be formed that

have distinct digits and are multiples of 12?

Any multiple of 12 should be a multiple of 4 and 3. First, let us look at the

constraint for a number being a multiple of 3. Sum of the digits should be a

multiple of 3. Sum of all numbers from 2 to 7 is 27. So, if we have to drop a digit

and still retain a multiple of 3, we should drop either 3 or 6.

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So, the possible 5 digits are 2, 4, 5, 6, 7 or 2, 3, 4, 5, 7.

When the digits are 2, 4, 5, 6, 7. the last two digits possible for the number to be a

multiple of 4 are 24, 64, 52, 72, 56, 76. For each of these combinations, there are 6

different numbers possible. So, with this set of 5 digits we can have 36 different

numbers

When the digits are 2, 3, 4, 5, 7. the last two digits possible for the number to be a

multiple of 4 are 32, 52, 72, 24. For each of these combinations, there are 6

different numbers possible. So, with this set of 5 digits we can have 24 different

numbers

Overall, there are 60 different 5-digit numbers possible

2. All numbers from 1 to 200 (in decimal system) are written in base 6 and base 7

systems. How many of the numbers will have a non-zero units digit in both base 6

and base 7 notations?

If a number written in base 6 ends with a zero, it should be a multiple of 6. In other

words, the question wants us to find all numbers from 1 to 200 that are not

multiples of 6 or 7. There are 33 multiples of 6 less than 201. There are 28

multiples of 7 less than 201. There are 4 multiples of 6 & 7 (or multiple of 42)

from 1 to 200.

So, total multiples of 6 or 7 less than 201 = 33 + 28 - 4 = 57. Number of numbers

with non-zero units digit = 200-57 = 143.

3. All numbers from 1 to 150 (in decimal system) are written in base 6 notation.

How many of these will not contain any zero?

Any multiple of 6 will end in a zero. There are 25 such numbers. Beyond this, we

can have zero as the middle digit of a 3-digit number. This will be the case for

numbers from 37-41, 73-77, 109-113 and 145-149. There are 20 such numbers.

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Overall, there are 45 numbers that have a zero in them.

4. How many factors of 1080 are perfect squares?

1080 = 2^3 * 3^3 * 5. For any perfect square, all the powers of the primes have to

be even numbers. So, if the factor is of the form 2^a * 3^b * 5^c. The values 'a' can

take are 0 and 2, b can take are 0 and 2, and c can take the value 0. Totally there

are 4 possibilities. 1, 4, 9, and 36.

1. Australia, in one of their matches, scored a total that is the lowest multiple of 11

with exactly 10 factors. Against who was this?

2^4 * 11 = 176. Australia scored this against Pakistan

2. In one of the matches, the top-scorer (who scored a century) for the team batting

first scored 3/8th the team's runs, but the team still did not cross 300. The HCF of

the team's score and the top-scorers score is a prime number. Which match was

this?

Top-scorer's score is a multiple of 3 greater than 100. So, it has to be a score like

34 * 3, 35 * 3, 36 * 3,..... The overall score is less than 300. So, the overall score

has to be lower than 38 * 8. So, the overall score has to be of the form 37 * 8, 36 *

8, 35 * 8,...

The HCF is a prime number, this implies that the top scorer scored 111, the team

scored 296. India vs. South Africa. Sachin Tendulkar century

3. A 'minnow' scored a total that can be made a perfect square if it is multiplied by

23. If they had scored 1 more, then their score multiplied by 13 would have yielded

a perfect square. What was their score? Bonus point, what match was this?

The score should have been 23 * 4 or 23 * 9, 23 * 9 = 207. 207 + 1 = 208, which is

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13 * 16.

India vs. Ireland. Ireland scored 207

4. In a particular match, team A batting first scored twice a perfect square. Team B

batting second, also scored twice a perfect square. The total score of both the teams

put together was also a perfect square. What match are we talking about here?

India 338, England 338 (169 * 2). Total 676 (26* 26)

5. Team scored a three digit score 'abc' in a match. a and b are prime numbers. a +

2b and a + 2c are also prime. b + 2a and b + 2c are also primes. a + c and b + c are

also primes. Two digit numbers 'ab' and 'ba' are both primes. Which match was

this?

a + c and b + c are also primes => a, b should be odd, c should be even

a,b cannot be 5 as ab is also prime. So, abh has to be 37.

c can only be zero.

abc = 370 (India vs. Bangladesh)

1. A number n! is written in base 6 and base 8 notation. Its base 6 representation ends

with 10 zeroes. Its base 8 representation ends with 7 zeroes. Find the smallest n

that satisfies these conditions. Also find the number of values of n that will satisfy

these conditions.

Base 6 representation ends with 10 zeroes, or the number is a multiple of 6^10. If

n! has to be a multiple of 6^10, it has to be a multiple of 3^10. The smallest

factorial that is a multiple of 3^10 is 24!. So, when n = 24, 25 or 26, n! will be a

multiple of 6^10 (but not 6^11).

Similarly, for the second part, we need to find n! such that it is a multiple of 2 ^ 21,

but not 2 ^ 24. When n = 24, n! is a multiple of 2^22. S0, when n = 24, 25, 26, 27,

n! will be a multiple of 2 ^ 21 but not 2 ^ 24.

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The smallest n that satisfies the above conditions is 24. n = 24, 25 or 26 will satisfy

the above conditions.

2. [x] is the greatest integer less than or equal to x. Find the number of positive

integers n such that [n/11] = [n/13].

This is a classic case of brute-force counting.

When n = 1, 2, 3....10, both values will be equal to 0. 10 possibilities

When n = 13,14, ....21, both values will be equal to 1. 9 possibilities



When n = 52, 53, 54, both values will be equal to 4. 3 possibilities

when n = 65, both will be equal to 5. 1 possibility

So, totally there are 1 + 3 + 5 + 7 + 9 +10 = 35 possibilities.

3. Positive numbers 1 to 55, inclusive are placed in 5 groups of 11 numbers each.

What is the maximum possible average of the medians of the 5 groups?

We need to maximise each median in order to have the overall maximum median

possible.

The highest possible median is 50 as they should be 5 numbers higher than the

median in the group of 11. So, if we have a set that has a, b, c, d, e, 50, 51, 52, 53,

54, 55, the median will be 50. In this set, it is best not to waste any high values on

a, b, c, d or e as these do not affect the median. So, a set that reads as 1, 2, 3, 4, 5,

50, 51, 52, 53, 54, 55 will also have median 50.

The next set can be 6, 7, 8, 9, 10, 44, 45, 46, 47, 48, 49. The median will be 44.

2. 1. Average of 6 distinct positive integers is 33. The median of the three largest

numbers is 43. What is the difference between the highest and lowest possible

median of the 6 numbers?

3. Let a, b, c, d, e, f be the six numbers in ascending order

4. e = 43; a + b + c + d + e + f = 198

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5. Median of 6 integers = (c +d)/2

6. (c +d)/2 is maximum when c and d are maximum since e = 43, dmax = 42 and cmax =

41

7. (c +d)/2 is minimum when c and d are minimum amin = 1, bmin = 2, cmin = 3 dmin = 4

8. Therefore, = (c +d)/2 min = 3.5

9. Max – min = 41.5 – 3.5 = 38

10.

11. 2. In class A, the ratio of boys to girls is 2:3. In class B the ratio of boys to girls

is 4 : 5. If the ratio of boys to girls in both classes put together is 3 : 4, what is

the ratio of number of girls in class A to number of girls in class B? 12. Let us assume class A has 2x boys and 3x girls, class B has 4y boys and 5y girls.

13. (2x + 4y)/(3x + 5y) = ¾

14. 8x + 16y = 9x + 15y

15. x = y

16. Reqd ratio = 3x/5y (since x = y) = 3/5

17. 3. N is an 80-digit positive integer (in the decimal scale). All digits except the

44th digit (from the left) are 2. If N is divisible by 13, find the 26th digit. 18. To begin with, the question should read "find the 44th digit".

19. Any number of the form abcabc is a multiple of 1001. 1001 is 7 * 11 * 13. So, any

number of the form abcabc is a multiple of 13.

20. So, a number comprising 42 2's would be a multiple of 13, so would a number

comprising 36 2's. So, in effect, we are left with a two digit number 2a, where a is

the 44th digit. 26 is a multiple of 13, so the 44th digit should be 6.

21. 22. 4. A page is torn from a novel. The sum of the remaining digits is 10000. What

is the sum of the two page-numbers on the torn page of this novel?

23. n(n+1)/2 should be nearby 10,000, so n(n+1) is somewhere near 20,000. So n

should be around sqrt(20,000) about 141. Try 142

24. 142*143 = 10153

25. 141 *142 = 10011

26. So, the missing pages are either 76 and 77 or 5 & 6

1) Pipe A, B and C are kept open and together fill a tank in t minutes. Pipe A is kept open

throughout, pipe B is kept open for the first 10 minutes and then closed. Two minutes

after pipe B is closed, pipe C is opened and is kept open till the tank is full. Each pipe

fills an equal share of the tank. Furthermore, it is known that if pipe A and B are kept

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open continuously, the tank would be filled completely in t minutes. Find t?

A is kept open for all t minutes and fills one-third the tank. Or, A should be able to fill the

entire tank in '3t' minutes.

A and B together can fill the tank completely in t minutes. A alone can fill it in 3t

minutes.

A and B together can fill 1/t of the tank in a minute. A alone can fill 1/3t of the tank in a

minute. So, in a minute, B can fill 1/t - 1/3t = 2/3t. Or, B takes 3t/2 minutes to fill an

entire tank

To fill one-third the tank, B will take t/2 minutes. B is kept open for t - 10 minutes.

t/2 = t - 10, t = 20 minutes.

A takes 60 mins to fill the entire tank, B takes 30 minutes to fill the entire tank. A is kept

open for all 20 minutes. B is kept open for 10 minutes.

C, which is kept open for 8 minutes also fills one-third the tank. Or, c alone can fill the

tank in 24 minutes.

2) Pipe A fills a tank at the rate of 100lit/min, Pipe B fills at the rate of 25 lit/min, pipe C

drains at the rate of 50 lit/min. The three pipes are kept open for one minute each, one

after the other. If the capacity of the tank is 7000 liters, how long will it take to fill the

tank if

i) A is kept open first, followed by B and then C

Each cycle of 3 minutes, 75 liters get filled. 100 + 25 - 50. So, after 3 minutes the tank

would have 75 liters

6 mins - 150 liters

9 mins - 225 liters

30 mins - 750 liters




In the 280th minute, pipe A would be open and it would fill the remaining 25 liters in 15

seconds. So, it would take 279 mins and 15 seconds to fill the tank.

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The most important thing in these type of questions is to think in terms of cycles till we

reach close to the required target and then think in simple steps.

ii) B first, followed by A, and then C

Each cycle of 3 minutes, 75 liters get filled. 25 + 100 - 50. So, after 3 minutes the tank

would have 75 liters

6 mins - 150 liters

9 mins - 225 liters





In the 280th minute, pipe B would be open and it would fill the remaining 25 liters in one

minutes. So, it would take 280 mins.

iii) B first, followed by C, and then A

In this case also, Each cycle of 3 minutes, 75 liters get filled. 25 -50 + 100. But there is a

small catch here. In the first set of 3 minutes, we would fill up to about 100 liters. After 1

minute, we would be at 25 liters, after 2 minutes, we would be at 0 liters and in the third

minute, the tank would be 100 liters full.

6 mins - 170 liters

9 mins - 250 liters





279 minutes - 7000 liters

So, it would take 279 mins and 15 seconds to fill the tank.

The most important thing in these type of questions is to think in terms of cycles till we

reach close to the required target and then think in simple steps.

3) 4 men and 6 women complete a task in 24 days. If the women are at least half as

efficient as the men, but not more efficient than the men, what is the range of the number

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of days for 6 women and 2 men to complete a task twice as difficult as this one?

4m and 6w finish in 24 days.

In one day, 4m + 6w = 1/24 of task

In these questions, just substitute extreme values to get the whole range

If a woman is half as efficient as man

4m + 3m = 1/24, 7m = 1/24, m = 1/168

6w + 2m = 3m + 2m = 5m, 5m will take 168/5 days = 33.6 days

If a woman is as efficient as a man

4m + 6w finish in 24 days

10m finish 1/24 of task in a day

6w + 2m = 8m, 8m will take 240/8 = 30 days to finish the task.

So, the range = 30 to 33.6 days. The new team will take 30 to 33.6 days to finish the task.

4) A fill pipe can fill a tank in 20 hours, a drain pipe can drain a tank in 30 hours. If a

system of n pipes (fill pipes and drain pipes put together) can fill the tank in exactly 5

hours, which of the following are possible values of n (More than one option could be

correct)?

1) 32 2) 54 3) 29 4) 40

3 fill pipes cancel out 2 drain pipes. Plus, you need an additional 4 fill pipes fill the tank

in 5 hours. so the answer has to be 5k + 4. Both 54 and 29 are possible.

1. a is x % of b, b is x% more than a. Find x?

a = bx,

b = a (1+x), substituting this in the previous equation, we get

x(x+1) = 1

x^2 + x -1 = 0

or x is approximately 0.62 or 62%

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2. A two digit number ab is 60% of x. The two-digit number formed by reversing the

digits of ab is 60% more than x. Find x?

10a + b = 0.6x

10b + a = 1.6x

Subtracting one from the other, we have

9b - 9a = x or x = 9( b-a)

x should be a multiple of 9.

10a + b = 3x/5, this implies that x should also be a multiple of 5.Or, x should be a

multiple of 45

x should be equal to 45, ab = 27, ba = 72

3. A is x % more than B. B is y% less than C. If A,B and C are positive and A is greater

than C, find the relation between x and y

a = (1 +x)b

b = c(1-y)

c = b /(1-y)

a is greater than c

Or, b (1+x) > b/(1-y)

1 + x > 1/(1-y)

(1+x)(1-y) > 1

1 + x -y - xy > 1

x > y (1+x)

y < x/(1+x)

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1. The median of n distinct numbers is greater than the average, does this mean that there

are more terms above the average than below it?

No.

If there are an odd number of terms, say, 2n+1, then the median is the middle term. And

if average is lesser than the middle term, there will at least be n +1 terms greater than the

average. So, there will be more terms above the average than below it.

However, this need not be the case when there are an even number of terms. When there

are 2n distinct terms, n of them will be greater than the median and n will be lesser than

the median. The average of these two terms can be such that there are n terms above the

average and n below it.

For instance, if the numbers are 0, 1, 7, 7.5. The median is 4, average is 3.875. Average is

less than the median. And there are more 2 numbers above the average and 2 below the

average.

2. In a sequence of 25 terms, can 20 terms be below the average? Can 20 terms be

between median and average?

Yes. We can have 24 zeroes and 1500 as the 25 numbers. In this case, there are 24

numbers below the average.

No. In a sequence of 25 numbers, 12 will be greater than or equal to the median and 12

will be lesser than or equal to the median. We cannot have 20 terms in between the

average and median

3. From 10 numbers, a,b,c,...j, all sets of 4 numbers are chosen and their averages

computed. Will the average of these averages be equal to the average of the average of

the 10 numbers?

Yes. All the terms appear the same number of times when we select them 4 at a time. So,

the average of averages will be equal to the overall average.

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1. Consider three classes A, B and C with 20, 30 and 50 students respectively. The

averages score in maths of students in class B is 16 more than that of those in class C and

the average score of those in class A is 2 more than the overall average of scores in A, B

and C. What is the difference between average of class C and class A

Let average in class C = x. Average of class B = x +16. Average of class A = y

y = 2 + (20y + 30(x+16) + 50x) /100

(y - 2) * 10 = 2y + 3x + 48 + 5x

10y -20 = 2y + 8x + 48

8y - 8x = 68

y - x = 8.5. This is the difference between average of class A and class C

2. Natural numbers 1 to 25 (both inclusive) are split into 5 groups of 5 numbers each. The

medians of these 5 groups are A, B, C, D and E. If the average of these medians is m,

what are the smallest and the largest values m can take?

Let us try to construct a scenario for getting the smallest value of the average. We need to

have the minimum value for each of the 5 medians.

Now, what is the lowest value the median of any of the 5 groups can take?

The median is the middle term among the 5 terms. So, it cannot be 1 or 2, but it can be 3.

If the set were 1, 2, 3, 4, 5, the median would be 3. If we choose a sub-group such as this,

the smallest median fro the next group will be 8. The next group could be 6, 7, 8, 9, 10.

The idea is to create sub-groups in such a way so that the second group can have a

smaller median than 8.

Now, even a sub-group that has the numbers 1, 2, 3, 24, 25 would have the median as 3.

So, if we want to keep the medians as small as possible, we should choose sub-groups

that have small numbers till the median, and then very large numbers. Because, these

large numbers do not affect the median, we will still have small numbers to choose from

for the next group. So, choose sub-groups such that the first 3 numbers are small, the

final two are as large as possible.

1, 2, 3, 24, 25

4, 5, 6, 22, 23

7, 8, 9, 20, 21

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10, 11, 12, 18, 19

13, 14, 15, 16, 17

would be an ideal set of groups. The medians would be 3, 6, 9, 12, 15, and the average of

the medians would be 9.

Similar set of groups can be found to find the highest value of the average. The medians

would be 23, 20, 17, 14, 11 and the average would be 17.

Q. Consider 4 numbers a, b, c and d. Ram figures that the smallest average of some three

of these four numbers is 30 and the largest average of some three of these 4 is 40. What is

the range of values the average of all 4 numbers can take?

We can assume a, b,c d are in ascending order (with the caveat that numbers can be equal

to each other)

a + b + c = 90

b + c + d = 120

We need to find the maximum and minimum value of a + b + c + d.

a + b + c + d = 120 + a. So, this will be minimum when a is minimum. Given a + b + c =

90. a is minimum when b + c is maximum. If b + c is maximum, d should be minimum.

Given that b + c + d = 120, the minimum value d can take is 40 as d cannot be less than b

or c. The highest value b +c can take is 80, when b = c= d = 40. When b = c = d = 40, a =

10. a + b + c + d = 130. Average = 32.5

Similarly, a + b + c + d = 90 + d. So, this will be maximum when d is maximum. Given b

+ c + d = 120. d is maximum when b + c is minimum. If b + c is minimum, a should be

maximum. Given that a + b + c = 90, the maximum value a can take is 30 as a cannot be

greater than b or c. The lowest value b +c can take is 60, when a = b= c = 30. When a = b

= c = 30, d = 60. a + b + c + d = 150. Average = 37.5

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So, the average has to range from 32.5 to 37.5

Q. The average of 5 distinct positive integers if 33. What are the maximum and minimum

possible values of the median of the 5 numbers if the average of the three largest numbers

within this set is 39?

Let the numbers be a, b, c, d, e in ascending order. a + b + c + d + e = 165. Average of

the three largest numbers is 39, so c + d + e = 117, or a + b = 48.

We need to find the maximum and minimum possible values of c.

For minimum value, a and b have to be minimum. a +b = 48. Let us assume a =23, b =25,

ca can be as low as 26.

23, 25, 26, 45, 46 is a possible sequence that satisfies the conditions.

For maximum value, we need d + e to be minimum as c + d + e = 117. we can have c =

38, d =39 and e =40. Or, the maximum value c can take = 38

23, 25, 38, 39, 40 is a possible sequence that satisfies the conditions specified

Q. Consider 5 distinct positive numbers a, b, c, d, and e. The average of these numbers is

k. If we remove b from this set, the average drops to m (m is less than k). Average of c, b,

d and e is K. We also know that c is less than d and e is less than k. The difference

between c and b is equal to the difference between e and d. Average of a, b, c and e is

greater than m. Write down a, d, c, d and e in ascending order.

Average of a, b, c, d and e is k, Average of b, c, d and e is also k, this implies that a = k

If we remove b, the average drops, this implies that b is higher than the average

e is less than k, or e is less than a. c is less than d.

e < a < b, c < d

Average of a, b, c and e is greater than average a, c, d and e. This tells us that d < b

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From this, we get that b is the largest number

b - c = d - e

b + e = c + d

a + b + c +d + e = 5a

Or, b + c + d + e = 4a

b + e = 2a, c + d = 2a. Or, e, a, b is an AP, c, a, d is an AP.

e is the smallest number (as b is the largest number)

Or, the ascending order should be e c a d b

1. Two friends A and B simultaneously start running around a circular track . They run in

the same direction. A travels at 6m/s and B runs at b m/s. If they cross each other at

exactly two points on the circular track and b is a natural number less than 30, how many

values can b take?

Let track length be equal to T. Time taken to meet for the first time = T / relative speed

= T/(6-b) or T/(b-6)

Time taken for a lap for A = T/6

Time taken for a lap for A = T/b

So, time taken to meet for the first time at the starting point = LCM (T/6, T/b) = T / HCF

(6,b)

Number of meeting points on the track = Time taken to meet at starting point/Time taken

for first meeting = Relative speed / HCF (6,b). For a more detailed discussion on this

have a look at the last few slides in this presentation.

So, in essence we have to find values for b such that 6-b/ HCF(6,b) = 2 or b-6/ HCF(6,b)

= 2

http://iimcat.blogspot.com/2010/11/races-basic-questions-solutions.html

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b = 2, 10, 18 satisfy this equation. So, there are three different values that b can take.

2. Three friends A, B and C decide to run around a circular track. They start at the same

time and run in the same direction. A is the quickest and when A finishes a lap, it is seen

that C is as much behind B as B is behind A. When A completes 3 laps, C is the exact

same position on the circular track as B was when A finished 1 lap. Find the ratio of the

speeds of A,B and C?

Let track length be equal to T. When a completes a lap, let us assume B has run a

distance of (t-d). At this time, C should have run a distance of (t-2d)

After three laps C would have traveled a distance of 3 * (t-2d) = 3t - 6d.

After 3 laps C is in the same position as B was at the end one lap. So, the position after

3t-6d should be the same as t-d. Or, C should be at a distance of d from the end of the lap.

C will have completed less than 3 laps (as he is slower than A), so he could have traveled

a distance of either t-d or 2t-d.

=> 3t-6d = t-d => 2t = 5d => d = 0.4t => The distances covered by A,B and C when A

completes a lap will be t, 0.6t and 0.2t respectively. Or, the ratio of their speeds is 5:3:1

In the second scenario, 3t-6d = 2t-d => t = 5d=> d = 0.2t => The distances covered by

A,B and C when A completes a lap will be t, 0.8t and 0.6t respectively. Or, the ratio of

their speeds is 5:4:3

The ratio of the speeds of A, B and C is either 5:3:1 or 5:4:3

Station X of length 900 meters has two station masters A and B. But as the station is not a

busy one, they are mostly jobless and decide to conduct an experiment. They stand at

either end of the station and decide to note the exact time when trains cross the

stationmasters. They synchronize their watches and proceed to either end of the station.

Two trains P and Q go past the station (neither train stops here), and after having taken

down their readings, the station masters sit down to have a chat

1. A: Train P entered the station at exactly 8:00:00

2. B: Train Q entered the station at exactly 8:00:10 (10 seconds past 8)

3. A: The last carriage of train P crossed me by at 8:00:20, and precisely two seconds

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after this, the engines of the two trains went past each other. (Engines are at the front of

the train)

4. B: The last carriage of train Q crossed me 22 seconds after the engine of P went past

me.

5. A: After the last carriage of train P crossed by me, it took 35 seconds for the engine of

train Q to cross me.

6. B: I got bored and I came here.

Now, let us try to jot down the points in a slightly different format and see if we can

make any inferences. Let us assume speed of train P = p, speed of train Q = q, length of

train P = L, length of train Q = M

Take statements 1 and 3

Train P crosses the stationmaster A entirely in 20 seconds (enters station at 8:00:00 and

last carriage passes at 8:00:20) => Length of train = 20p => L = 20p

Statement 5 tells us that engine of train Q crosses station master A at 8:00:55. Train Q

enters the station at 8:00:10, so the train takes 45 seconds to cross the entire station. Or, it

takes 45 seconds to travel 900 meters => Speed to train Q, q = 20m/s.

Statement 3 tells us that the two trains cross each other at 8:00:22. This implies train P

has traveled for 22 seconds since entering the station and train Q has traveled for 12

seconds since entering the station before they cross each other. The cumulative distance

traveled by the two trains should be equal to the length of the station = 900m.

=> 900 = 22p + 12q

q = 20m/s => p = 30m/s.

So, train P_ takes 900/p = 30 seconds to cross the station. So, engine of train P will cross

stationmaster B at 8:00:30.

Statement 4 states that the last carriage of Q went past 22 seconds after the engine of P

went by. Or, the last carriage of Q went by at 8:00:52.

Engine of train Q went by at 8:00:10, last carriage went by at 8:00:52, or train Q took 42

seconds to cross station master B. Train Q travels at 20m/s. => Length of train Q = 840m

Now, to the questions

1. What is the length of train Q?

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Train Q is 840m long

2. At what time do the rear ends of the two trains cross each other?

The engines cross each other at 8:00:22. The relative speed of the two trains = 20+30 =

50m/s. The relative distance traveled by two trains from the time the engines cross each

other to the times the rear ends cross each other = Sum of the two lengths = 600 + 840

=1440. Time taken = 1440/50 = 28.8 seconds past 8:00:22, or at 8:00:50.8 seconds.

3. How far from station master A do the rear ends of the two trains cross each other?

At 8:00:50.8, P would have traveled 50.8 * 30m/s post entering the station. Or, train P

would have traveled 1524m. The rear end of train P would be at a point 1524-600m =

924m from stationmaster A (or 24 meters from stationmaster B and outside the station)

4. You are told that the two trains enter the station at the same times mentioned and the

length of the two trains are unchanged. Furthermore, train P continues to travel at the

same speed (as computed above). At what minimum speed should train Q travel such that

the rear ends of the two trains cross each other at a point within the length of the

platform?

Rear end of train P crosses the station completely at 8:00:50. (Train P takes 30 seconds to

travel the station and 20 seconds to travel a distance equal to its length). Train Q should

have traveled 840m by this time. => Train Q should travel 840 within 40 seconds.

Or, minimum speed of Q = 840/40 = 21m/s.

1. Two friends A and B leave City P and City Q simultaneously and travel towards Q

and P at constant speeds. They meet at a point in between the two cities and then

proceed to their respective destinations in 54 minutes and 24 minutes respectively.

How long did B take to cover the entire journey between City Q and City P?

Let us assume Car A travels at a speed of a and Car B travels at a speed of b.

Further, let us assume that they meet after t minutes.

Distance traveled by car A before meeting car B = a*t. Likewise distance traveled

by car B before meeting car A = b * t.

Distance traveled by car A after meeting car B = a *54. Distance traveled by car B

after meeting car A = 24* b

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Distance traveled by car A after crossing car B = distance traveled by car B before

crossing car A (and vice versa)

=> at = 54b -------- 1

and bt = 24a -------- 2

Multiplying equations 1 and 2

we have ab * t2 = 54 * 24 * ab

=> t2 = 54 * 24 => t = 36.

So, both cars would have traveled 36 minutes prior to crossing each other. Or, B

would have taken 36 + 24 = 60 minutes to travel the whole distance.

2. Car A trails car B by 50 meters. Car B travels at 45km/hr. Car C travels from the

opposite direction at 54km/hr. Car C is at a distance of 220 meters from Car B. If

car A decides to overtake Car B before cars B and C cross each other, what is the

minimum speed at which car A must travel?

To begin with, let us ignore car A. Car B and car C travel in opposite directions.

Their relative speed = Sum of the two speeds = 45 + 54 kmph. = 99kmph. = 99 *

5/18 m/s = 55/2 m/s = 27.5m/s.

The relative distance = 220m. So, time they will take to cross each other =

220/27.5 = 8 seconds.

Now, car A has to overtake car B within 8 seconds. The relative distance = 50m

=> Relative speed should be at least 50/8m/s. = 6.25m/s

= 6.25 * 18/5 kmph = 22.5kmph.

Car B travels at 45kmph, so car A should travel at at least 45 + 22.5 = 67.5kmph.

2. 1. City A to City B is a downstream journey on a stream which flows at a speed of

5km/hr. Boats P and Q run a shuttle service between the two cities that are 300

kms apart. Boat P, which starts from City A has a still-water speed of 25km/hr,

while boat Q, which starts from city B at the same time has a still-water speed of

15km/hr. When will the two boats meet for the first time? (this part is easy) When

and where will they meet for the second time?

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When boat P travels downstream, it will effectively have a speed to 30kmph.

Likewise, Q will have an effective speed of 10kmph. The relative speed = 40kmph.

So, the two boats will meet for the first time after 300/40 hours (Distance/relative

speed) = 7.5 hours (Actually, for this part we do not need the speed of the stream)

The second part is more interesting, because the speed of the boats change when

they change direction. Boat P is quicker, so it will reach the destination sooner.

Boat P will reach City B in 10 hours (300/30). When boat P reaches city B, boat Q

will be at a point 100kms from city B.

After 10 hours, both P and Q will be traveling upstream

P's speed = 20km/hr

Q's speed = 10 km/hr => Relative speed = 10kmph

Q is ahead of P by 100 kms

P will catch up with Q after 10 more hours (Relative Distance/relative speed -

100/10).

So, P and Q will meet after 20 hours at a point 200 kms from city B

3.

4. 2. Cities M and N are 600km apart. Bus A starts from city M towards N at 9AM

and bus B starts from city N towards M at the same time. Bus A travels the first

one-third of the distance at a speed of 40kmph, the second one-third at 50kmph and

the third one-third at 60km/hr. Bus B travels the first one-third of the total time

taken at a speed of 40kmph, the second one-third at 50kmph and the third one-third

at 60km/hr. When and where will the two buses cross each other?

5.

Bus A

Travels 200km at 40kmph

the next 200km @ 50kmph and

the final 200km @ 60kmph

So, Bus A will be at a distance of 200km from city M after 5 hours, and at a

distance of 400km after 9 hours, and reach N after 12 hours and 20 mins

Bus B

Travels at an overall average speed of 50kmph, so will take 12 hours for the entire

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trip

So, Bus B will travel

160kms in the first 4 hours

200 kms in the next 4

and 240 in the final 4

So, both buses cross each other when they are in their middle legs.

After 5 hours, bus A will be at a position 200kms from city M. At the same time,

bus B will be at a distance 210kms from city N (4*40+50).

The distance between them will be 190kms (600-200-210). Relative speed = Sum

of the the two speeds = 50+50 = 100 kmph.

Time taken = 190/100 = 1.9 hours. = 1 hour and 54 minutes. So, the two buses will

meet after 6 hours and 54 minutes. Bus B will have travelled 210 + 95 = 305 kms.

So, the two buses will meet at a point that is 305 kms from City N and 295 kms

from city A.

1. A 4-digit number of the form aabb is a perfect square. What is the value of a-b?

A number of the form aabb has to be a multiple of 11. So, it is the square of either 11 or

22 or 33 or...so on up to 99.

88^2 = 7744. This is the only solution possible. Most of these trial and error questions

need to be narrowed down a little bit before we can look for the solution. That narrowing

down is critical. In this case, we should look for multiples of 11.

2. LCM of three numbers is equal to 1080; HCF of the three numbers is equal to 2. How

many such triplets are possible?

HCF = 2. Let the numbers be 2x, 2y and 2z. HCF of (x,y,z) =1

LCM = 2xyz. => xyz = 540

540 = 22 * 33 * 5

x = 22 , y = 33, z = 5

x = 22 * 33, y = 5, z = 1

x = 22 * 5, y = 33, z = 1

x = 33 * 5, y = 22, z = 1

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x = 22 * 33 * 5, y=1, z=1

These are the basic combinations. The first 4 yield 6 ordered triplets each, the last one

yields 3 ordered triplets.

So, there are 5 unordered triplets and 27 ordered triplets

3. N leaves a remainder of 4 when divided by 33, what are the possible remainders when

N is divided by 55?

The LCM of 33 and 55 is 165. This is the starting point. If a number leaves a remainder

of 4 when divided by 33, it can leave remainders 4, 37, 70, 103, and 136.

Or the number can be of the form 165n +4, or 165n + 37 or 165n + 70,103 or 136

When divided by 55, the possible remainders are 4, 37, 15, 48 and 26

Fermat's Little Theorem and Euler's Phi function

Fermat's Little Theorem states that if p is prime then a^p - a is a multiple of p. In other

words,

a ^ (p-1) = 1 mod p, whenever a is not a multiple of p

The proof for this theorem is interesting. Take any number a that is co-prime with p.

Now, a will correspond to some r (mod) p, where r lies between 1 and p-1

Now, take the numbers a, 2a, 3a, 4a, 5a,...(p-1)a...All these numbers will be co-prime

with p (as p is prime) and all will leave remainders from 1 to p-1. Let us assume they

leave remainders r, r2,r3,r4, …rp-1. Now, none of these remainders will be equal to 0.

Importantly, no two of these remainders can be equal.

(This is a critical result, which is established below)

Suppose r3 were equal to r5. Then, 5a -3a would be a multiple of p => 2a is a multiple of p,

which is impossible.

This implies r, r2,r3,r4, …rp-1 should correspond to 1,2,3,...p-1 in some order.

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Now, when we multiply a, 2a, 3a, 4a,...(p-1)a in modular arithmetic terms, we should get

a remainder of 1*2*3*...(p-1)

Or, a p-1 * 1*2*3*...(p-1) = 1*2*3*...(p-1) mod p

Let us say 1*2*3*...(p-1) = X

Or, a p-1 * X = X mod p

Or, X (a p-1 -1) = 0 mod p

Now, X cannot be 0 mod p, so, a ^ (p-1) has to be 1 mod p, which is Fermat's Little

Theorem

Now, for Euler's Phi function and Theorem.

Euler's Phi function states that

For two numbers m,n that are coprime (HCF of m,n =1)

m phi(n) = 1 mod n, (m phi(n) leaves a remainder of 1 when divided by n)

Now, m and n are co-prime numbers. The possible remainders that m can have when

divided by n are numbers from 0 to n-1 that are co-prime to n. A set that has phi(n)

elements.

Now, let the elements in that set be R1 R2,R3,R4, …Rp set of possible remainders that m

can leave when divided by n. This implies that p = phi(n).

Now, let us assume that m leaves a remainder R on division by n, R belongs to the above

mentioned set.

Let us take the numbers R1 * m, R2 * m, R3 * m,R4 * m, …Rp * m. Now, none of these

would correspond to o mod n. Importantly, no two of these can be equal mod n either.

Because if R4 * m and R7 * m were equal mod n, then m *( R4 - R7) would be a multiple of

n, which is impossible.

Therefore, the numbers R1 * m, R2 * m, R3 * m,R4 * m, …Rp * m should correspond to the

numbers R1 , R2 , R3 ,R4, …Rp mod n in some order.

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Now, if we take the product of the p numbers R1 * m, R2 * m, R3 * m,R4 * m, …Rp * m.

This will be equal to R1 * R2 * R3 * R4* …* Rp mod n.

Let R1 * R2 * R3 * R4* …* Rp = Y

mp Y = Y mod n

Y (mp-1) = 0 mod n

Or, mp = 1 mod n

m ^ phi(n) = 1 mod n, which is Euler's Phi Function.

Brilliant Theorem. Beautiful implications. But, I dont think CAT will have a question

based on this theorem.

If we divide a number N by divisor p and get remainder r, we can write N = pq + r. In

modulus arithmetic, we say that is N = r (mod) p.

r can take values from 0 to p-1. If the divisor is 12, the remainder can be anywhere from

0 to 11. If the remainder comes out as 13, this is the same as 1. If we compute the

remainder as -5, this is the same as +7

Now, modulus is consistent for addition, subtraction & multiplication. What we mean by

this is

If a = x mod p and

b = y mod p

Then a+b = (x+y) mod p

a-b = (x-y) mod p and

ab = xy mod p

If the divisor is 12, the remainder can be anywhere from 0 to 11. If the remainder comes

out as 13, this is the same as 1. If we compute the remainder as -5, this is the same as +7

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Now, let us take this discussion on mod a little further. Now, let us assume two numbers

a and b that are co-prime to each other. Further let us assume a = r mod b.

a= kb + r

HCF (a,b)= 1 => HCF (bk+r,b) = 1, this implies that HCF (r,b) =1. Or, in other words r

and b have to be co-prime.

Let us think about this property with some examples. Let us take b = 12. Any number that

leaves a remainder, say, 4 when divided by 12 can be written as a = 12n + 4 = 4(3n +1) ,

or a is a multiple of 4 => a cannot be co-prime with 12.

So, if we know a is co-prime with 12, then we can say that the only remainders possible

when a is divided by 12 are 1, 5, 7,11 - Numbers that are co-prime with 12.

Or, the number of possible remainders of a when divided by b, when a,b are

coprime = phi (b). . A concept which we used to solve question no 3 in this set .

Further, if we say a,b are coprime. Any power of a will be co-prime with b. And So, a ^n

will leave a remainder within the set of numbers that are lower than b and co-prime to b.

Let us call this set Euler Set. Let us name the complement of this set the non-Euler set.

So, for 12, the Euler set will contain the elements 1,5,7,11. The non-Euler set will contain

0, 2,3,4,6,8,9,10.

Any two numbers with mod within the Euler set will give a product with a

modulus within the Euler set.

If multiply n numbers together, and even one of the numbers leaves a remainder in

the non-Euler set, the overall product will leave a remainder in the non-Euler set.

We have discussed some of the basic properties of mod in this post. Let us have a re-look

Euler's Phi Function.

Euler's phi function is an important property in Number Theory. But before we go in any

detail into this topic, let me clarify that it is very unlikely that a CAT question will

http://iimcat.blogspot.com/2010/11/number-theory-eulers-phi-function.html

http://iimcat.blogspot.com/2010/11/number-theory-remainders.html

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require students to know Euler's Phi function. Before, we go into Euler's Phi function, it

is probably good to have a look at the mod function.

or phi(n) is defined as the number of natural numbers less than or equal to n and are

coprime to n.

phi(4) = 2 (The numbers 1 and 3)

phi(12) = 4 (The numbers 1, 5, 7 and 11)

phi(13) = 12 (All numbers below 13)

In general phi(p) = p-1 when is a prime.

Generalising further, when N = paqbrc phi(N) = N (1-1/p) * (1-1/q) * (1-1/r) (The proof for this is intuitive enough. Keep

eliminating all numbers that are not coprime from 1 to N-1)

Now, Euler's phi function states this

For two numbers m,n that are coprime (HCF of m,n =1)

m phi(n) = 1 mod n, (m phi(n) leaves a remainder of 1 when divided by n)

m phi(n) – 1 is a multiple of n

When we apply it in a scenario where n is prime, we get

m p-1 = 1 mod p

m p-1 – 1 is a multiple of n

This last observation is also called Fermat's Little Theorem . In many ways, Euler's phi

function is an extension of Fermat's Little Theorem.

Excellent theorem, fairly clear implications whenever a question on remainders is asked.

For instance if a question states What is the remainder when 100^56 is divided by 29, we

can straight away see that the answer is 1. Having said that, I do not think that a CAT

question will require students to know Euler's phi function and properties. Any question

that gets simplified using Euler's phi function will have a simpler alternate solution as

well. Even if you solve using Euler's phi function, kindly look for the alternate method.

http://en.wikipedia.org/wiki/Fermat%27s_little_theorem

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1. How many pairs of integers (x,y) exist such that the product of x, y and HCF (x,y) =

1080?

We need to find ordered pairs (x, y) such that xy*HCF(x, y) = 1080.

Let x = ha and y = hb where h = HCF(x, y) => HCF(a, b) = 1.

So h^3(ab) = 1080 = (2^3)(3^3)(5).

We need to write 1080 as a product of a perfect cube and another number.

Four cases:

I h = 1, ab = 1080 and b are co-prime. We gave 4 pairs of 8 ordered pairs (1,1080), (8,

135), (27,40) and (5,216) (Essentially we are finding co-prime a,b such that a*b = 1080)

II h = 2, We need to find number of ways of writing (3^3) * (5) as a product of two

coprime numbers. This can be done in two ways - 1 and (3^3) * (5) , (3^3) and (5)

number of pairs = 2, number of ordered pairs = 4

III - h = 3, number of pairs = 2, number of ordered pairs = 4

IV - h = 6, number of pairs = 1, number of ordered pairs = 2

Hence total pairs of (x, y) = 9, total number of ordered pairs = 18.

The pairs are (1,1080), (8, 135), (27,40), (5,216), (2,270), (10, 54), (3,120), (24,15) and

(6,30),

2. Find the smallest number that leaves a remainder of 4 on division by5, 5 on division by

6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

LCM(5, 6, 7, 8, 9) - 1 = 2519.

3. There are three numbers a,b, c such that HCF (a,b) = l, HCF(b,c) =m and HCF (c,a) =

n. HCF(l,m) = HCF(l,n) = HCF(n,m) =1. Find LCM of a,b,c. (The answer can be "This

cannot be determined").

a is a multiple of l and n. Also HCF (l,n) =1; => a has to be a multiple of ln, similarly b

has to be a multiple of lm and c has to be a multiple of mn.

We can assume, a = lnx, b = lmy, c = mnz.

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Now given that HCF(a, b) = l, that means HCF(nx, my) = 1. This implies HCF(x, y) = 1

and HCF(m, x) = HCF(n, y) = 1.

Similarly it can also be shown that HCF(y, z) = HCF(z, x) = 1 and others also.

So in general it can be written any two of the set {l, m, n, x, y, z} are co-prime.

Now LCM(a, b, c) = LCM(lnx, lmy, mnz) = lmnxyz = abc/lmn.

Quiet obviously, it is a reasonable assumption that a question in CAT will not be as tough

as the last one here. However, it is a good question to get an idea of the properties of

LCM and HCF.

1. How many pairs of positive integers x,y exist such that HCF of x,y = 35 and sum of x

and y = 1085?

Let HCF of (x,y) be h. Then we can write x = h*a and y = h*b. Furthermore, note that

HCF (a,b) = 1. This is a very important property. One that seems obvious when it is

mentioned but a property a number of people overlook.

So, we can write x = 35a; y = 35b

x + y = 1085 => 35( a + b) = 1085. => (a+b) = 31. We need to find pairs of coprime

integers that add up to 31. (Another way of looking at it is to find out integers less than

31 those are coprime with it or phi(31) as one of the replies had mentioned. More on this

wonderful function in another post).

Since 31 is prime. All pairs of integers that add up to 31 will be coprime to each other.

Or, there are totally 15 pairs that satisfy this condition.

2. How many pairs of positive integers x,y exist such that HCF(x,y) + LCM (x,y) = 91?

This has become one of my favourite questions. Had not thought much about it when I

had written down the question - but came to realise that there are perhaps many more

interesting questions that can be created in this genre.

http://en.wikipedia.org/wiki/Euler%27s_totient_function

90

Again let us write x = h * a; y = h * b,

a and b are coprime. So, LCM of (x,y) = h*a*b

So, in essence h + h*a*b = 91. Or h(ab + 1) = 91

Now, 91 can be written as 1*91 or 7*13

Or, we can have HCF as 1, LCM as 90 - There are 4 pairs of numbers like this (2,45),

(9,10), (1,90) and (5,18)

We can have HCF as 7, ab +1 as 13 => ab =12 => 1*12 or 4*3

Or, the pairs of numbers are (7,84) or (21,28)

The third option is when HCF = 13, ab+1 = 7 => ab=6

Or (a,b) can be either (1,6) or (2,3)

The pairs possible are (13,78) and (26,39)

There are totally 8 options possible - (2,45), (9,10), (1,90), (5,18), (7,84), (21,28), (13,78)

and (26,39)

3. Sum of two numbers x,y = 1050. What is the maximum value of the HCF between x

and y?

x=525 y =525 works best.

If the question states x,y have to be distinct, then the best solution would be x=350 y

=700, HCF = 350

1. What are the last two digits of the number 745 ?

The last two digits of 71 are 07

The last two digits of 72are 49



The last two digits of powers of 7 go in a cycle - 07,49,43,01

So, the last two digits of 745 are 07

2. What is the remainder when we divide 390 + 590 by 34?

390 + 590 can be written as (32)45 + ( 52)45

91

= (9)45 + (25)45

Any number of the form an + bn is a multiple of (a + b) whenever n is odd

So (9)45 + (25)45 is a multiple of 9 +25 = 34

So, the remainder when we divide (32)45 + ( 52)45 by 34 is equal to 0

3. N2 leaves a remainder of 1 when divided by 24. What are the possible remainders we

can get if we divide N by 12?

This again is a question that we need to solve by trial and error. Clearly, N is an odd

number. So, the remainder when we divide N by 24 has to be odd.

If the remainder when we divide N by 24 = 1, then N2 also has a remainder of 1. we can

also see that if the remainder when we divide N by 24 is -1, then N2 a remainder of 1

2 has a remainder of 2 2 2

When remainder when we divid 2 has a remainder of 1

Or, the possible remainders when we divide N by 24 are 1, 5, 7, 11, 13, 17, 19, 23

Or, the possible remainders when we divide N by 12 are 1, 5, 7, 11

A prime number p greater than 100 leaves a remainder q on division by 28. How many

values can q take?

q can be 1.

If q =2, number would be of the form 28n + 2 which is a multiple of 2

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Similarly, when q =4, number would be of the form 28n + 4 which is again a multiple of

2. Any number of the form 28n + an even number will be a multiple of 2

When q =7, number would be of the form 28n + 7 which is a multiple of 7

So, the only remainders possible are remainders that share no factors with 28. Or

numbers that are co-prime to 28.

There is a formula for this and a shorter way of finding the number of numbers co-prime

to a given natural number. A more detailed discussion on this is provided here and here.

1,3,5,9,11,13,15,17,19,23,25 and 27. q can take 12 different values.

Qn2:

How many positive integers are there from 0 to 1000 that leave a remainder of 3 on

division by 7 and a remainder of 2 on division by 4?

Number should be of the form 7n + 3 and 4m +2

The LCM of 7 and 4 is 28. So, let us see what are the possible remainders when we

divide this number by 28

A number of the form 7n + 3 can be written as 28K + 3 or 28k + 10 or 28+ 17 or 28k

+24.

A number of the form 4m + 2 can be written as 28l +2, 28l + 6, 28l + 10, 28l + 14, 28l +

18, 28l + 22, 28l + 26,

For a detailed discussion on how we get to this, look at this post.

Within these, the only common term is 28K + 10.

The numbers in this sequence are 10, 38, 66.....990.

We still need to figure out how many numbers are there in this sequence. We are going in

steps of 28, so let us see if we can write these numbers in terms of 28p + r

10 = 28 * 0 + 10

38 = 28 * 1 + 10

66 = 28 * 2 + 10

94 = 28 * 3 + 10

…..

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990 = 28 * 35 + 10

There are 36 numbers in this sequence.

Another way of looking at this question is to spot the number of common terms in two

APs

{3,10,17,24 ,... } and {2,6,10,14,...}

Question 1:

Three numbers leave remainders of 43, 47 and 49 on division by N. The sum of the three

numbers leaves a remainder 9 on division by N. What are the values N can take?

Question 2.

A number leaves a remainder 3 on division by 14, and leaves a remainder k on division

by 35. How many possible values can k take?

Correct Answers:

Question 1: 65 and 130

Question 2: 5 different values

Explanatory Answers

Qn 1: Three numbers leave remainders of 43, 47 and 49 on division by N. The sum of the

three numbers leaves a remainder 9 on division by N. What are the values N can take?

This question is based on some very basic and very important remainder properties.

When the sum of two numbers is divided by the same divisor, the remainder should be

equal to the sum of the two remainders. As long as the divisor remains the same,

remainders are consistent for addition, subtraction and multiplication. In other words, we

can add three numbers and then compute the remainders, or just add the three remainders.

The equivalent rule applies for multiplication and subtraction also.

In this question, the sum of the three numbers should leave a remainder 43 + 47 + 49 =

139. Or, it should be of the form N * k + 139, where K is an integer.

However, it leaves a remainder 9, or it is of the form N * m + 9

N * k + 139 = N *m + 9

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N * (m-k) = 130.

Or, N should be a factor of 130. Since the remainders left on division by N are 43, 47 and

49, N should be greater than 49.

The only factors of 130 that are greater than 49 are 65 and 130. So, N can take 2 values –

65 or 130.

Qn 2: A number leaves a remainder 3 on division by 14, and leaves a remainder k on

division by 35. How many possible values can k take?

Let us have a look at the theory for this question as well. For instance, let us assume a

number N leaves a remainder of 3 on division by 8. What would be the remainder when

number N is divided by 24?

N/8 remainder = 3

N/24 remainder = ?

Let us look at Numbers that leave remainder 3 on division by 8

3, 11, 19, 27, 35, 43 ……

For these numbers, remainders when divided by 24 are

3, 11, 19, 3, 11, 19 ……

Possible remainders are 3, 11 or 19

Alternative approach

N/8 remainder = 3

N = 8q + 3

q can be in one of 3 forms

3p

3p + 1

3p + 2

N = 8(3p) + 3 or

8(3p + 1) + 3 or

8(3p + 2) + 3

24p + 3 or

24p + 11 or

95

24p + 19

N/24 possible remainders are 3, 11, 19

Why did we choose to write q as 3p, 3p + 1 or 3p + 2?

8 x 3 = 24, this is why we chose 3p, 3p +1, 3p + 2

So, if we are given that remainder on dividing N by 8, then there will be a set of

possibilities for the remainder of division of N by 24 (or any multiple of 8)

Let us look at the opposite also. Say, we know the remainder of division of N by 42 is 11,

what should be the remainder when N is divided by 7?

N/42 remainder = 11

N/7 remainder =?

N / 42 remainder = 11

N = 42q + 11

42q + 11 divided by 7

42q leaves no remainder

11/7 remainder = 4

So, if we are given that remainder on dividing N by 42, then we can find the remainder of

dividing N by 7 (or any factor of 42)

Now, let us address the question

A number leaves a remainder of 3 on division by 14, or it can be written as 14n + 3

On division by 70, the possible remainders can be 3, 17 (3 +14), 31 (3 + 28), 45 (3 + 42),

or 59 (3 + 56). The number can be of the form

70n + 3

70n + 17

70n + 31

70n + 45

70n + 59

Now, we need to divide this number by 35

96

70n + 3 divided by 35, the remainder will be 3





On division by 35, the possible remainders are 3, 17, 31, 10 or 24. There are 5 possible

remainders

Question

1. Given N is a positive integer less than 31, how many values can n take if (n+1) is a

factor of n! ?

Correct Answer:

18 values

Explanatory Answer:

The best starting point for this question is to do some trial and error.

3 is not a factor of 2!



6 is a factor of 5!


8 is a factor of 7!

The first thing we see is that n+1 cannot be prime. If (n +1) were prime, it cannot be a

factor of n!.

So, we can eliminate all primes.

Now, let us think of all numbers where (n+1) is not prime. In this instance, we should be

able to write (n+1) as a * b where a,b are not 1 and (n+1). So, (a,b) will lie in the set { 1,

2, 3, .....n} or, a* b will be a factor of (n + 1)!

So, for any composite (n+1), (n+1) will always be a factor of n! {Is there any exception?}

For any prime number (n +1), (n+1) will never be a factor of n!

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The above rule works well even for all the examples we have seen, except when (n+1) =

4. 4 = 2 * 2; So, 4 is not a factor of 3!. But this is the only exception.

Counting on from here, we can see that n can take values 5, 7, 8, 9, 11, 13, 14,15, 17, 19,

20, 21, 23, 24, 25, 26, 27, 29. Essentially, all numbers where N+1 is greater than 4 and is

not prime will feature in this list. If N+1 is not prime, we should be able to write is as a

product of 2 numbers less than N+1. This will feature in n!. This question is just a

different way of asking one to count primes (and then account for the exception of 4)

n can 18 different values.

Now, if we know that there are 25 prime numbers less than 100, can you answer the

following question - Given N is a positive integer less than 101, how many values can n

take if (n+1) is a factor of n!?

1. How many values can natural number n take, if n! is a multiple of 76 but not 79?

The smallest factorial that will be a multiple of 7 is 7!

14! will be a multiple of 72

Extending this logic, 42! will be a multiple of 76

However, 49! will be a multiple of 78 as 49 (7 * 7) will contribute two 7s to the factorial.

(This is a standard question whenever factorials are discussed). Extending beyond this,

56! will be a multiple of 79

In general for any natural number n,

n! will be a multiple of [n/7] + [n/49] + [n/343] + …..

where [x] is the greatest integer less than or equal to x. A more detailed discussion of this

is available on this link

So, we see than 42! is a multiple of 76. We also see that 56! is the smallest factorial that is

a multiple of 79. So, n can take values { 42, 43, 44, 45, ……55}

There are 14 values that n can take.

2. How many values can natural number n take, if n! is a multiple of 220 but not 320?

The highest power of 2 that will divide n! = [n/2] + [n/4] + [n/8] + [n/16]….. and so on.

So, let us try to find the smallest n such that n! is a multiple of 220,

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If n = 10, the highest power of 2 that will divide n! = [10/2] + [10/4] + [10/8] = 5 + 2 + 1

= 8

If n = 20, the highest power of 2 that will divide n! = [20/2] + [20/4] + [20/8] + [20/16] =

10 + 5 + 2 + 1 = 18

If n = 24, the highest power of 2 that will divide n! = [24/2] + [24/4] + [24/8] + [24/16] =

12 + 6 + 3 + 1 = 22

{Here we can also see that each successive number is just the quotient of dividing the

previous number by 2. As in, [12/2] = 6, [6/2] = 3, [3/2] = 1. This is a further shortcut one

can use.}

So the lowest number of n such that n! is a multiple of 2^20 is 24

Now, moving on to finding n! that is a multiple of 3. The highest power of 3 that will

divide n! = [n/3] + [n/9] + [n/27] + [n/81] and so on,

When n = 20, the highest power of 3 that can divide 20! = [20/3] + {6/3] = 6 + 2 = 8

When n = 35, the highest power of 3 that can divide 35! = [35/3] + {11/3] + [3/3] = 11 +

3 +1 = 15

When n = 45, the highest power of 3 that can divide 45! = [45/3] + [15/3] + [5/3] = 15 +

5 +1 = 21

The lowest number n such that n! is a multiple of 3^20 is 45.

When n takes values from 24 to 44, n! will be a multiple of 2^20 and not 3^20. n can take

21 values totally.

Question

1. Find the least number n such that no factorial has n trailing zeroes, or n+1 trailing

zeroes or n+2 trailing zeroes?

Correct Answer

153

Explanatory Answer

The previous question includes a detailed discussion on how to find the number of

trailing zeroes of n!, for any natural number n.

We see that 24! has [24/5] = 4 zeroes

25! ends with [25/5] + [25/25] = 6 zeroes. There is no natural number m such that m! has

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exactly 5 zeroes.

Similarly, we see that 49! ends with [49/5] + [49/25] = 10 zeroes, whereas 50! ends with

[50/5] + [50/25] = 12 zeroes. No factorial ends with 11 zeroes.

So, any time we have a multiple of 25, we 'skip' a zero. This is because a multiple of 25

adds two zeroes to the factorial.

Extrapolating this, we can see that 125 might actually 'skip' two zeroes. 124! ends with

[124/5] + [124/25] = 24 + 4 = 28 zeros, whereas 125! has [125/5] + [125/25] + [125/125]

= 25 + 5 +1 = 31 zeros. there is no factorial with 29 or 30 zeros.

In order to jump three zeros, think about what we need to look at. Every multiple of 25

gives us one 'skipped' zero. Every multiple of 125 gives us two 'skipped' zeroes.

In order to have three skipped zeroes, we need to look at 624! and 625!

624! has [624/5] + [624/25] + [624/125] = 124 + 24 + 4 = 152 zeros

625! has [625/5] + [625/25] + [625/125] + [625/125] = 125 + 25 + 5 + 1 = 156 zeros

There is no factorial with 153, 154 or 155 zeros. Or the least value of n such that no

factorial ends with n, (n+1) or (n+2) zeroes is 153

1. How many trailing zeroes (zeroes at the end of the number) does 60! have?

To start with, the number of trailing zeroes in the decimal representation of a number =

highest power of 10 that can divide the number.

For instance,

3600 = 36 * 102

45000 = 45 * 103

In order to approach this question, let us first see the smallest factorial that ends in a zero.

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

100

Now, 5! ends in a zero as we have get a product of 10 when we compute 1 * 2 * 3 * 4 * 5

10 is 2*5, so we get a factor of 10 every time we get a 2 and a 5 in the factorial.

So, 5! has 1 zero. The factorial that ends with 2 zeroes is 10!

15! has 3 zeroes.

20! has 4 zeroes and so on.

An extra zero is created every time a 2 and 5 combine. Every even number gives a two,

while every fifth number gives us a 5.

Now, the critical point here is that since every even number contributes at least a 2 to the

factorial, 2 occurs way more frequently than 5. So, in order to find the highest power of

10 that can divide a number, we need to count the highest power of 5 that can divide that

number. We do not need to count the number of 2’s in the system as there will be more

than 2’s than 5’s in any factorial.

Now, every multiple of 5 will add a zero to the factorial. 1 * 2 * 3 *…..59 * 60 has

twelve multiples of 5. So, it looks like 60! will end in 12 zeroes. But we need to make

one more adjustment here.

25 is 52, so 25 alone will contribute two 5’s, and therefore add two zeroes to the system.

Likewise, any multiple of 25 will contribute an additional zero.

So, 20! has 4 zeroes, 25! has 6 zeroes.

60! will have [60/5] zeroes arising due to the multiples of and an additional [60/25] due

to the presence of 25 and 50. {We retain only the integer component of 60/25 as the

decimal part has no value}

So, 60! will end with 12 + 2 zeros. = 14 zeros.

In general, any n! will end with [n/5] + [n/25] + [n/125] + [n/625] …… zeroes.

Generalizing further, in case we want to find the highest power of 3 that divides n!, this is

nothing but [n/3] + [n/9] + [n/27] + [n/81] ……

The highest power of 7 that divides n! is [n/7] + [n/49] + [n/343] ……

101

In case of a composite number, we need to break into the constituent primes and compute

the highest power that divides the number.

For instance, if we want to find the largest power of 15 that divides n!, this will be driven

by the highest powers of 3 and 5 that divide n!. Similar to the scenario we saw with

trailing zeroes, we can observe that there will definitely be at least as many 3’s than 5’s in

any factorial. So, the highest power of 15 that divides n! is simply [n/5] + [n/25] +

[n/125] + [n/625] ……

2. What is the highest power of 12 that divides 54! ?

12 = 22 * 3, so we need to count the highest power of 2 and highest power of 3 that will

divide 54! and then we can use this to find the highest power of 12.

The method to find highest powers of 2 and 3 are similar to the one outlined in the

previous question.

Highest power of 2 that divides 54! = [54/2] + [54/4] + [54/8] + [54/16] + [54/32] = 27 +

13 + 6 + 3 + 1 = 50

Highest power of 3 that divides 54! = [54/3] + [54/9] + [54/27] = 18 + 6 + 2 = 26

Or 54! is a multiple of 250 * 326. Importantly, these are the highest powers of 2 and 3 that

divide 54!.

22 * 3 = 12. We need to see what is the highest power of 22 * 3 that we can accommodate

within 54!

In other words, what is the highest n such that (22*3)n can be accommodated within 250*

326

Let us try some numbers, say, 10, 20, 30

(22 *3)10 = 220 * 310, this is within 250 * 326

(22 *3)20 = 240 * 320, this is within 250 * 326

(22 *3)30 = 260 * 330, this is not within 250 * 326

The highest number possible for n is 25.

102

(22 *3)25 = 250 * 325, this is within 250 * 326, but (22 *3)26 = 252 * 326, this is not within 250

* 326

So, 54! can be said to be a multiple of (22 * 3)25. Or, the highest power of 12 that can

divide 54! is 25.

Note: For most numbers, we should be able to find the limiting prime. As in, to find the

highest power of 10, we need to count 5s. For the highest power of 6, we count 3s. For

15, we count 5s. For 21, we count 7’s. However, for 12, the limiting prime could be 2 or

3, so we need to check both primes and then verify this.

Question

Let the cost of 3 apples and 4 oranges be Rs 21. If Anjli can buy at most 4 apples and 3

oranges with Rs 20 and no more fruit, then the maximum amount that could be left with

Anjli will be about?

First set of replies

· Ans is 2.50 /- . correct me if i am wrong

· If each apple cost 33paisa and oranges 5 rupees then Anjali can buy 3 apples and 4 apples

for 21then 4 apples and 3 oranges will cost 16.2 which means that she can have 3.8 rs.

For such problems u need to assign maximum values to more quantities and then should

proceed

· say let me follow with yo approach! even then ! i can assume value 5.1 for b? right ? and

4*(5.1) + 3*(0.2) = 21. so here b = 5.1 and a=0.2 and 4(0.2) + 3(5.1) will be 16.1 and the

answer would be 3.9 right ?? Kindly any one reply to this !

· Orange-Rs. 4, Apple-Rs. 1.66. Balance- Rs. 1.33, Hope You got it

Question recap (this mail sent by yours truly, must confess that I fell for the trap)

Let the cost of 3 apples and 4 oranges be Rs 21. If Anjli can buy at most 4 apples and 3

oranges with Rs 20 and no more fruit, then the maximum amount that could be left with

Anjli will be about?

I would go for pricing apples at 0, price per orange would be Rs. 5.25 Cost of 4 apples

and 3 oranges = 3* 5.25 + 0 = Rs.15.75, which leaves us with Rs. 4.25

Of course, this assumes that the price of apple cannot be negative :-)

103

Correct Solution - This is a beautiful solution.

all are missing one point that with 20 she can not buy any more fruit ..means the money

left is less than price (minimum) of any og the two friuts...otherwise she can buy more

friuts, so max amount of money left can be just higher than cost of chaper fruit.

Now 4x+3y =21

also 3x+4y+left money =20

3x+4y+y=20(leftmoney=cost of cheaper fruit, y will be chaper as on exchange of

qunatity total value decreases or 4x+3y>3x+4y..........x>y)

4x+3y =21

3x+5y =20

solving this y=17/11=1.545 which is the answer

This is why this is a beautiful question (final reply from question-provider)

@sony, thank you so much, thats exactly what we were missing

a)1 b)1.25 c)2 d)2.25 e)1.5

these are the options given,

from the options given i was ignoring the costs that include denominations other than

normally what we see, but i'm getting the answer as .75(with apple=2 and orange=3.75)

from your solution the answer is 1.5, i guess thats the correct one from the options.

Explanations

1. How many factors of the number 2^8 * 3^6 * 5^4 * 10^5 are multiples of 120?

The prime factorization of 28 * 36 * 54 * 105 is 213 * 36 * 59

For any of these factors questions, start with the prime factorization. Remember that the

formulae for number of factors, sum of factors, are all linked to prime factorization.

120 can be prime-factorized as 23 * 3 * 5

104

All factors of 213 * 36 * 59 that can be written as multiples of 120 will be of the form 23 *

3 * 5 * K

213 * 36 * 59 = 23 * 3 * 5 * K => K = 210 * 35 * 58

The number of factors of N that are multiples of 120 is identical to the number of factors

of K.

Number of factors of K = (10+1) (5+1) * (8+1) = 11 * 6 * 9 = 594


Any factor of 213 * 36 * 59 will be of the form 2p * 3q * 5r . When we are trying to find the

number of factors without any constraints, we see that

p can take values 0, 1, 2, 3, ….13 – 14 values

q can take values 0, 1, 2, …6 – 7 values

r can take values 0, 1, 2, 3,….9 – 10 values

So, the total number of factors will be 14 * 7 * 10.

This is just a rehash of our formula (a+1) ( b + 1) ( c +1)

In this scenario we are looking for factors of 213 * 36 * 59 that are multiples of 120. These

will also have to be of the form 2p * 3q * 5r. But as 120 is 23 * 3 * 5, the set of values p, q,

r can take are limited.

p can take values 3, ….13 – 11 values

q can take values 1, 2, …6 – 6 values

r can take values 1, 2, 3,….9 – 9 values

So, the total number of factors that are multiples of 120 will be 11 * 6 * 9 = 594.

2. Number N = 2^6 * 5^5 * 7^6 * 10^7; how many factors of N are even numbers?

The prime-factorization of 26 * 55 * 76 * 107 is 213 * 512 * 76

The total number of factors of N = 14*13*7

105

We need to find the total number of even factors. For this, let us find the total number of

odd factors and then subtract this from the total number of factors. Any odd factor will

have to be a combination of powers of only 5 and 7.

Total number of odd factors of 213 * 512 * 76 = (12+1) * (6 + 1) = 13 * 7

Total number of factors = (13+1) * (12+1) * (6 + 1)

Total number of even factors = 14 * 13 * 7 - 13 * 7

Number of even factors = 13 * 13 * 7 = 1183


Any factor of 213 * 512 * 76 will be of the form 2p * 5q * 7r .

Any even factor of 213 * 512 * 76 will also be of the same form, except that p cannot be

zero in this case

p can take values 1, 2, 3, ….13 – 13 values

q can take values 0,1, 2, …12 – 13 values

r can take values 0, 1, 2, 3,….5 – 7 values

So, the total number of even factors be 13 * 13 * 7 = 1183

1. Numbers A, B, C and D have 16, 28, 30 and 27 factors. Which of these could be a

perfect cube?

Any number of the form paqbrc will have (a+1) (b+1)(c+1) factors, where p, q, r are prime.

In order for the number to be a perfect cube a, b, c will have to be multiples of 3.

We can assume that a = 3m, b = 3n, c = 3l

This tells us that the number of factors will have to be of the form (3n+1) *(3m+1)

*(3l+1). In other words (a +1), (b + 1) and (c + 1) all leave a remainder of 1 on division

by 3. So, the product of these three numbers should also leave a remainder of 1 on

division by 3. Of the four numbers provided, 16 and 28 can be written in this form, the

other two cannot.

106

So, a perfect cube can have 16 or 28 factors. Now, let us think about what kind of

numbers will have 16 factors.

A number of the form p15 or q3r3 will have exactly 16 factors. Both are perfect cubes.

Note that there are other prime factorizations possible that can have exactly 16 factors.

But these two forms are perfect cubes, which is what we are interested in

Similarly, a number of the form p27 or q3r6 will have 28 factors. Both are perfect cubes.

Given the number of factors, one should be able to write down the basic prime

factorization forms that could lead to these many factors. This is a critical skillset for

tackling some of the tougher questions in CAT.

2. If a three digit number ‘abc’ has 3 factors, how many factors does the 6-digit number

‘abcabc’ have?

‘abc’ has exactly 3 factors, so ‘abc’ should be square of a prime number. (This is an

important inference, please remember this).

Any number of the form paqbrc will have (a+1) (b+1)(c+1) factors, where p, q, r are prime.

So, if a number has 3 factors, its prime factorization has to be p2

‘abcabc’ = ‘abc’ *1001 or abc * 7*11*13 (again, this is a critical idea to remember)

Now, ‘abc’ has to be square of a prime number. It can be either 121 or 169 (square of

either 11 or 13) or it can be the square of some other prime number

When abc = 121 or 169, then ‘abcabc’ is of the form p3q1r1 1, which should have 4*2*2 =

16 factors

When ‘abc’ = square of any other prime number (say 172 which is 289) , then ‘abcabc’ is

of the form p1q1r1s2 , which should have 2*2*2*3 = 24 factors

So, ‘abcabc’ will have either 16 factors or 24 factors

Qn1 :How many numbers are there less than 100 that cannot be written as a multiple of a

perfect square?

107

To begin with, all prime numbers will be part of this list. There are 25 primes less than

100. (That is a nugget that can come in handy)

Apart from this, any number that can be written as a product of two or more primes will

be there on this list. That is, any number of the form pq, or pqr, or pqrs will be there on

this list (where p, q, r, s are primes). A number of the form pnq cannot be a part of this list

if n is greater than 1, as then the number will be a multiple of p2.

This is a brute-force question.

First let us think of all multiples of 2 * prime number. This includes 2*3, 2*5, 2*7, 2 * 11

all the way up to 2*47 (14 numbers)

The, we move on to all numbers of the type 3 * prime number 3*5, 3*7 all the way up to

3*31 (9 numbers)

Then, all numbers of the type 5 * prime number – 5*7, 5*11, 5*13, 5*17, 5*19 (5

numbers),

Then, all numbers of the type 7 * prime number and then 7*11, 7*13 (2 numbers).

There are no numbers of the form 11 * prime number which have not been counted

earlier.

Post this, we need to count all numbers of the form p*q*r, where p, q, r are all prime.

In this list, we have 2*3*5, 2*3*7, 2*3*11, 2*3*13 and 2*5*7. Adding 1 to this list, we

get totally 36 different composite numbers.

Along with the 25 prime numbers, we get 61 numbers that cannot be written as a product

of a perfect square greater than 1

Alternative Method

There is another method of solving this question.

108

We can list all multiples of perfect squares (without repeating any number) and subtract

this from 99

4 - there are 24 multiples of 4 { 4, 8, 12, …96}

9 - There are 11 multiples, 2 are common with 4 (36 and 72), so let us add 9 new

numbers to the list{ 9, 18, 27, ….99}

16 - 0 new multiples

25 - 3 new multiples { 25, 50, 75

36 – 0 new ones

49 – 2 { 49, 98}

64 - 0

81 - 0

So, total multiples of perfect squares are 38. There are 99 numbers totally. So, there are

61 numbers that are not multiples of perfect squares

This is a difficult and time-consuming question. But a question that once solved, helps

practice brute-force counting. Another takeaway is the fact that there are 25 primes less

than 100. There is a function called pi(x) that gives the number of primes less than or

equal to x. pi(10) = 4, pi(100) = 25

4. Find the smallest number that has exactly 18 factors?



Now, the number we are looking for has 18 factors. It can comprise one prime, two

primes or three primes.

Now, 18 can be written as 1 * 18 or 3 * 6 or 9 * 2 or 2 * 3 * 3

If we take the underlying prime factorization of N to be paqb, then it can be of the form

p1q8 or p2 q5

If we take the underlying prime factorization of N to be pa, then it can be of the form

p17

If we take the underlying prime factorization of N to be paqbrc, then it can be of the form

p1q1r2

So, N can be of the form p17, p2q5, p1q8 or p1q2r2

109

Importantly, these are the only possible prime factorizations that can result in a number

having 18 factors.

Now, let us think of the smallest possible number in each scenario

p17 - Smallest number = 217

p2q5 – 32 * 25

p1q8 – 31 * 28

p1q1r2 – 51 * 31 * 22

The smallest of these numbers is 51 * 31 * 22 = 180

Questions

1. A number N^2 has 15 factors. How many factors can N have?

2. If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many

factors does the 6-digit number ‘abcabc’ have?

Correct Answer

Question 1: 6 or 8 factors

Question 2: 16 factors

Explanatory Answer

Qn: A number N^2 has 15 factors. How many factors can N have?



N2 has 15 factors.

Now, 15 can be written as 1 * 15 or 3 * 5.

If we take the underlying prime factorization of N2 to be paqb, then it should have (a + 1)

(b+1) factors. So, N can be of the form

p14 or p2q4

p14 will have (14+1) = 15 factors

p2q4 will have (2+1) * (4 +1) = 15 factors.

Importantly, these are the only two possible prime factorizations that can result in a

number having 15 factors.

110

Qn: If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many factors

does the 6-digit number ‘abcabc’ have?

To start with ‘abcabc’ = ‘abc’ *1001 or abc * 7*11*13 (This is a critical idea to

remember)

‘abc’ has only two factors. Or, ‘abc’ has to be prime. Only a prime number can have

exactly two factors. (This is in fact the definition of a prime number)

So, ‘abcabc’ is a number like 101101 or 103103.

’abcabc’ can be broken as ‘abc’ * 7 * 11 * 13. Or, a p * 7 * 11 * 13 where p is a prime.

As we have already seen, any number of the form paqbrc will have (a+1) (b+1)(c+1)

factors, where p, q, r are prime.

So, p * 7 * 11 * 13 will have = (1+1)*(1+1)*(1+1)*(1+1) = 16 factors

Education

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