01.Number System.pdf

1.NUMBER SYSTEM

I. NUMERAL : In Hindu Arabic system, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7,

8, 9 called digits to represent any number.

A group of digits, denoting a number is called a numeral.

We represent a number, say 689745132 as shown below :

Ten

Crores

(108)

Crores

(107)

Ten Lacs

(Millions)

(106)

Lacs

(105)

Ten

Thousands

(104)

Thousands

(103)

Hundreds

(102)

Tens

(101)

Units

(100)

6 8 9 7 4 5 1 3 2

We read it as : 'Sixty-eight crores, ninety-seven lacs, forty-five thousand, one

hundred and thirty-two'.

II.PLACE VALUE OR LOCAL VALUE OF A DIGIT IN A NUMERAL:

In the above numeral:

Place value of 2 is (2 x 1) = 2; Place value of 3 is (3 x 10) = 30;

Place value of 1 is (1 x 100) = 100 and so on.

Place value of 6 is 6 x 108 = 600000000

III.FACE VALUE: The face value of a digit in a numeral is the value of the digit

itself at whatever place it may be. In the above numeral, the face value of 2 is 2;

the face value of 3 is 3 and so on.

IV.TYPES OF NUMBERS

1.Natural Numbers : Counting numbers 1, 2, 3, 4, 5,..... are called natural

numbers.

2.Whole Numbers : All counting numbers together with zero form the set of whole

numbers. Thus,

(i) 0 is the only whole number which is not a natural number.

(ii) Every natural number is a whole number.

3.Integers : All natural numbers, 0 and negatives of counting numbers i.e.,

{, - 3 , - 2 , - 1 , 0, 1, 2, 3,..} together form the set of integers.

(i) Positive Integers : {1, 2, 3, 4, ..} is the set of all positive integers.

(ii) Negative Integers : {- 1, - 2, - 3,..} is the set of all negative integers.

(iii) Non-Positive and Non-Negative Integers : 0 is neither positive nor

negative.

So, {0, 1, 2, 3,.} represents the set of non-negative integers, while

{0, - 1 , - 2 , - 3 , ..} represents the set of non-positive integers.

4. Even Numbers: A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8,

10, etc.

5. Odd Numbers: A number not divisible by 2 is called an odd number. e.g., 1, 3, 5,

7, 9, 11, etc.

Points to Remember

The sum of any number of even numbers is always even. The sum of odd number of odd numbers (i.e., the sum of 3 odd numbers,

the sum of 5 odd numbers, etc.)' is always odd.

The sum of even number of odd numbers (i.e., the sum of 2 odd numbers, the sum of 4odd numbers, etc.) is always even.

The product of any number of odd numbers is always odd. The product of any number of numbers where there is at least one even

number is even Every even number greater than 4 can be expressed as a sum of two odd

prime numbers.

Difference between the squares of two consecutive numbers is always an odd number (it is the sum of those two numbers).

6. Perfect Numbers

A number is said to be a perfect number if the sum of ALL its factors

excluding itself (but including 1) is equal to the number itself.

For example, 6 is a perfect number because the factors of 6, i.e., 1, 2

and 3 add up to the number 6 itself.

Other examples of perfect numbers are 28, 496, 8128. etc.

7. Prime Numbers: A number greater than 1 is called a prime number, if it has

exactly two factors, namely 1 and the number itself.

Prime numbers upto 100 are : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,

43,47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

There is no general formula that can give prime numbers. Every prime

number greater than 3 can be written in the form of (6k + 1) or (6k - 1) where k is

an integer. For the proof of this,

Prime numbers Greater than 100 : Let p be a given number greater than 100.

To find out whether it is prime or not, we use the following method :

Find a whole number nearly greater than the square root of p. Let k > *jp.

Test whether p is divisible by any prime number less than k. If yes, then p is not

prime. Otherwise, p is prime.

e.g, We have to find whether 191 is a prime number or not. Now, 14 >191.

Prime numbers less than 14 are 2, 3, 5, 7, 11, 13.

191 is not divisible by any of them. So, 191 is a prime number.

8. Composite Numbers: Numbers greater than 1 which are not prime, are known as

composite numbers, e.g., 4, 6, 8, 9, 10, 12.

Facts:

(i) 1 is neither prime nor composite.

(ii) 2 is the only even number which is prime.

(iii) There are 25 prime numbers between 1 and 100.

(iv) When a three digit number is reversed & the difference of these two numbers

is taken, the middle number is always 9 & the sum of the other two numbers is

always 9

(v) All numbers may be real or complex

(vi) If the sum of the factors of any particular number other than itself is equal to

that particular number, then the number is said as perfect number.

(vii) When multiplication of consecutive numbers is divided with 2, it always gives

the remainder of 0.

9. Co-primes: Two numbers a and b are said to be co-primes, if their H.C.F. is 1. e.g.,

(2, 3), (4, 5), (7, 9), (8, 11), etc. are co-primes,

10. Relative Primes

Two numbers are said to be relative primes or co-primes if they do not

have any common factor other than 1. For example, the numbers 15 and 16

do not have any common factors and hence they are relative primes. Please

note that none of the two numbers may individually be prime and still they

can be relative primes. Unity is a relative prime to all numbers.

11. Multiples

If one number is divisible exactly by a second number, then the first

number is said to be a multiple of the second number. For example, 15 is a

multiple of 5; 24 is a multiple of 4.

12. Factors

If one number divides a second number exactly, then the first number

is said to be a factor of the second number.

For example, 5 is a factor of 15; 3 is a factor of 18.

Factors are also called sub-multiples or divisors.

13. Ramanujan's number The number 1729 is called Ramanujan's number. It is the only smallest number that can be expressed as a sum of two cubes in

two different ways 1729 = 123 + 13 = 103 + 93

14. Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144. ... is a fibonacci series. Starting from the third number, every number is the sum of the two

immediately preceding numbers.

15.Factorial The factorial of a whole number N is the continued product of all numbers

from 1 to N. It is denoted by N! or lN. Ex: 5! = 5 4 3 2 1 = 120

16. Fractional Numbers Numbers which have numerator & denominator (2/3)

i) Proper Fraction Number in which numerator is greater than denominator

ii) Improper Fraction Number in which denominator is greater than numerator

iii) Mixed Fraction Reason why mixed fraction has come in is to convert improper

fraction to a proper one since all fractions should be proper fractions.

iv) Rational Numbers Numbers where the decimals of their simplification

terminates at some point. Example 1/2, 3/2

v) Irrational Numbers Numbers where the decimals of their simplification do not

terminates. Example 2, , 5/3

vi) Real Numbers All the above seen numbers are real numbers

vii) Complex Numbers Numbers which are imaginary are complex numbers and are represented as I

V.TESTS OF DIVISIBILITY

Number Rule Examble

2 A number is divisible by 2, if its unit's digit is any of 0,

2, 4, 6, 8.

1986

3 A number is divisible by 3, if the sum of its digits is

divisible by 3.

9123

4 A number is divisible by 4 if the number formed with

its last two digits is divisible by 4.

178564

5 A number is. divisible by 5 if its last digit is 5 or zero 100, 2005

6 A number is divisible by 6 if it is divisible both by 2

and 3

12, 24

7 (i) Subtract two times the unit digit from the remaining

number. If it is divisible by 7, then the number is also

divisible by 7.

ii)Group the number into three from the right hand

side and find the sum of odd groups separately. The

difference between the sums should be divisible by 7.

1729, 4093992

8 A number is divisible by 8, if the number formed by the

last 3 digits of the number is divisible by 8.

3816, 14328,

18864

9 A number is divisible by 9 if the sum of its digits is a

multiple of 9.

60732

10 A number is divisible by 10, if it ends with 0. 1980, 19870

11 If the sum of the alternate digits is the same or they

differ by the sum at the digits in the even places in the

number should be equal to zero or a multiple of 11.

785345, 132

13 i) Add four times the unit digit to the remaining

number

ii) Group the number into three from the right hand

side and find the sum of odd groups separately. The

difference between the sums should be divisible by 13

1274

12, 14,

15

Divisibility by numbers like 12, 14, 15 can be checked

out by taking factors of the number which are

relatively prime and checking the divisibility of the

given number by each of the factors. For example, a

number is divisible by 12 if it is divisible both by'3 and

4.

16 if the number formed by the last4 digits is divisible by 7957536

16.

17 Subtract five times the unit digit from the remaining

number

187

19 Add two times the unit digit to the remaining number. 665, 969

23 Add seven times the unit digit to the remaining

number.

24 A given number is divisible by 24, if it is divisible by

both3 and 8.

40 Number should be divisible by both 5 and 8



VI. MULTIPLICATION BY DISTRIBUTIVE LAW:

(i) a x (b + c) = a x b + a x c (ii) a x (b-c) = a x (b-a) x c.

Ex.(i) 567958 x 99999 = 567958 x (100000 - 1)

= 567958 x 100000 - 567958 x 1 = (56795800000 - 567958) = 56795232042. (ii) 978 x

184 + 978 x 816 = 978 x (184 + 816) = 978 x 1000 = 978000.

VII. BASIC FORMULAE

Binomial Theorem: (x + y)n = n xn + n xn-1.y + n xn-2.y2 + .. + n yn

1. (a + b)2 = a2 + b2 + 2ab 2. (a - b)2 = a2 + b2 - 2ab

3. (a + b)2 - (a - b)2 = 4ab 4. (a + b)2 + (a - b)2 = 2 (a2 + b2)

5. (a2 - b2) = (a + b) (a - b) 6. (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)

7. (a3 + b3) = (a +b) (a2 - ab + b2) 8. (a3 - b3) = (a - b) (a2 + ab + b2)

9. (a3 + b3 + c3 -3abc) = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)

10. If a + b + c = 0, then a3 + b3 + c3 = 3abc.

SOME IMPORTANT RESULTS:

(i) Sum of first n, Natural numbers is (1 + 2 + 3 +. + n) =n(n+1)/2

(ii) Sum of the square of the first n Natural numbers (l2+22+32+..+n2)=

n(n+1)(2n+1)/6

(iii) Sum of the cubes of first n Natural number is (13 + 23 + 33 + ... + n3) =n2(n+1)2

(iv) Sum of the first n odd numbers, 1+3+5++n = n

(v) Sum of the first n even numbers, 2+4+6++n = n(n+1)

Surds and Indices:

Characteristics of square roots of numbers

1. If a square number ends in 9, its square root is a number ending in3 or 7.


3. If a square number ends in 5, its square root is a number ending in5



6. If a square number ends in 0, its square root is a number ending in 0.

VIII. DIVISION ALGORITHM

If we divide a given number by another number, then:

Dividend = (Divisor x Quotient) + Remainder

(i) (xn - an ) is divisible by (x - a) for all values of n.

(ii) (xn - an) is divisible by (x + a) for all even values of n.

(ii) (xn + an) is divisible by (x + a) for all odd values of n.

XI. FINDING NUMBER OF ZEROES IN THE GIVEN FACTORIAL

No. of Zeroes in 1i=1

!=5

n

i

xx ,

Where n= i when Numerator of x Denominator 5i

Example: Find the no. of zeroes in 165!

Here, x = 165

To find n: When i = 1, 165

5 5ix

, When i = 2, 165

5 25ix

When i = 3, 165

5 125ix

, When i = 4, 165

5 625ix

Hence n = 4

So, No. of Zeroes in 165! = 4 1

i=1

165165!=

5i

165 165 165

5 25 125 = 40 Zeroes

X. FINDING LARGEST / MAXIMUM POWER OF A NUMBER IN A GIVEN

FACTORIAL

Step a) Let the factor be x & the number to which maximum power is to be found

be n.

Step b) Divide x with n & take the quotient.

Step c) Divide that quotient with n & take its quotient.

Step d) Repeat the same till the quotient value becomes lesser than n

Step e) Sum up all obtained quotient values to find the maximum power

Example: Largest power of 2 in 30! is?

Step a) x = 30 & n = 2

Step b) 30/2 = 15

Step c) 15/2 = 7

Step d) 7/2 =3, 3/2 = 1 (1 < 2, so terminate here)

Step e) 15 + 7 + 3 + 1 = 26

XI. HIERARCHY OF ARITHMETIC OPERATIONS

To simplify arithmetic expressions, which involve various operations like

brackets, multiplication, addition, etc. a particular sequence of the operations has

to be followed.

For example, 2 + 3 x 4 has to be calculated by multiplying 3 with 4 and the

result 12 added to 2 to give the final result of 14 (you should not add 2 to 3 first to

take the result 5 and multiply this 5 by 4 to give the final result as 20). This is

because in 'arithmetic operations, multiplication should be done first before

addition is taken up.

The hierarchy of arithmetic operations are given by a rule called VBODMAS

rule. The operations have to be carried out in the order, in which they appear in the

word VBODMAS, where different letters of the word VBODMAS stand for the

following operations:

V-Vinculum

B- Brackets

O- Of

D- Division

M- Multiplication

A - Addition

S- Subtraction

There are four types of brackets:

(i) Vinculum: This is represented by a bar on the top of the numbers. For example, 2

+ 3 - 4 + 3; Here, the figures under the vinculum have to calculated as 4 + 3 first

and the "minus" sign before 4 is applicable to 7.Thus the given expression is

equal to 2 + 3 - 7 which is equal to -2.

(ii) Simple Brackets: These are represented by ( )

(iii)Curly Brackets: These are represented by { }

(iv) Square Brackets: These are represented by [ ]

After brackets is 0 in the BODMAS rule standing for "of' which means

multiplication. For example, 1/2 of 4 will be equal to 1/2 x 4 which is equal to 2.

After 0, the next operation is D standing for division. This is followed by M

standing for multiplication. After Multiplication, A standing for addition will be

performed. Then, S standing for subtraction is performed.

XII. FINDING UNIT DIGIT OF THE GIVEN NUMBER

Method 1

Let the number be x & p be the power value.

If the unit digit of x is 0, 1, 5 or 6, then the unit digit of xp = 0, 1, 5, 6 respectively

If the unit digit of x is 2, 3, 7 or 8, then divide p with 4 & take the remainder (r). If r

>0, then unit digit of xp = 2r, 3r, 7r or 8r. If r =0, then unit digit of xp =6, 1, 1 or 6

respectively.

If the unit digit is 4 or 9 and p is odd, then unit digit of xp =4 or 9 respectively & if p

is even, then unit digit of xp = 6 or 1 respectively.

Example: Unit place value of 235315=?

Here unit digit of x is 3, so p should be divided with 4. Here p = 15.

Remainder of 15/4 = 3

3r = 33 = 27

Unit place value = 7

Method 2

Example: Unit place value of 235315 =?

Unit Place value of 35 = 3 & hence 315 = 33 = 3 * 3 * 3 = 9 * 3 = 7

Unit place value =7

Similarly for other numbers such as 2, 7 & 8, when power is raised to 5, it gives the

same number as unit place. So it is not necessary to repeat the first step for every

problem.

UNIT PLACE CHART:

Number 1 2 3 4 5 6 7 8 9

Power of 1 1 2 3 4 5 6 7 8 9

Power of 2 1 4 9 6 5 6 9 4 1

Power of 3 1 8 7 4 5 6 3 2 9

Power of 4 1 6 1 6 5 6 1 6 1

Power of 5 1 2 3 4 5 6 7 8 9

XIII. FINDING NUMBER OF PRIME FACTORS/ NO OF DEVISORS

Let n = ap*bq*cr

i. No. of prime factors / No. of divisors = (p + 1) (q + 1) (r + 1)

Example: No. of prime factors in the expression 56 is?

56 = 2 * 2 * 2 * 7 = 2 * 7 (1, 2, 4, 7, 8, 14, 28, 56)

(3+1)* (1+1) = 8

ii. No of odd factor: Add all odd no+1 and multiple.

Ex: 56, odd factor is 1 & 7.. so take only 7 power value add 1.

iii. No of even factor =Total - odd factor.

Ex. 56 (2, 4, 8, 14, 28, 56)

The Sum of Factors/ Divisors = 1 1

1

pa

a

1 1

1

rb

b

Also sum of odd factor is take only odd numbers and even factor is total odd.

XIV. RECURRING DECIMAL TO A FRACTION

To convert recurring decimal (Obtained from irrational fraction) to a fraction,

the decimal should be considered as x first.

Case a) Let x = 0.333333 ------------------- (1) Here 3 (single digit) is getting repeated.

So multiply 10 on both the sides 10 x = 3.33333 ----------- (2) Subtract (1) from (2) 9x = 3 x = 1/3

Case b) Let x = 0.636363 ------------------- (1) Here 63 (two digits) is getting repeated.

So multiply 100 on both the sides 100 x = 63.6363 ----------- (2) Subtract (1) from (2) 99x = 63 x = 7/11

Case c) Let x = 0.16666. 100x = 16.666. (Here multiplying with 100 because two numbers in recurring decimal)

10x = 1.666 (Here multiplying with 10 because we need to eliminate the decimal value)

Subtracting above two equations we get, 90x = 15 x = 1/6

SOLVED PROBLEMS

1. 1+4+9+16+25+36+49+64+81+100 =?

1 + 2 + 3 + + 10 = [n (n+1) (2n+1)]/6 = [10 * 11 * 21]/6 = 385

2. What are the smallest numbers that must be added to 803642 in order to obtain

a multiple of 9 and 11 respectively?

Sum of the digits = 23

4 is to be added to 23 so that the number will be divisible by 9.

Sum of alternative numbers, 15 8 = 7 is to be added so that the no. will be divisible by 11.

Hence, (4, 7)

3. How many times should the key of a typewriter be pressed in order to type

number 300 pages of a book?

1 to 9 = 9 [How to get 9? 9-1+1] * 1 = 9

10 to 99 = 90 [How to get 90? 99-10+1] * 2 = 180

100 to 300 = 201 [How to get 201? 300-100+1] * 3 = 603

792

792 times

4. Find the unit place of 7 * 8.

Therefore, 7 = 7= 3 (power/4 and find reminder, then num to the power rem.) 8 = 8^1=8

3*8= 4

5. On dividing 15968 by a certain number, the quotient is 89 and the remainder is

37. Find the divisor.

Divisor = (Dividend Remainder) / Quotient = (15968 37) / 89 = 179

6. Find the remainder when 2^31 is divided by 5.

210 = 1024

210 * 210 * 210= ..4 * ..4 * ..4 = ..4

Therefore, 4 * 2 = 8

When any number which ends with 8 is divided by 5, it gives the remainder of 3.

(Or)

Find the power of 2 and divide by 5 where u get rem as 1, then write the no based on that power value.. 231 => ( 22=4 divide by 5 = rem 4 or -1, then -1* -1= 1 so 24 /5 rem =1)

(24) 7.2 3 = 1. 8=8 divide by 5 rem is 3.

7. {[2n+4 2(2)n ] / 2 * 2n+3} + 2-3 =? [[2n+4 2(2) n]/ [2 * 2n+3]] + 2-3 [2n*24 21*2n]/ [2 * 2n * 23] +2-3 {2n [24 - 2]} / 2 * 2n * 23 + [1/23] (1/23){[(24 2)/2] +1} = 8/8 = 1

8. What is the largest 4 digit no. that leaves the remainder of 6 when divided by 15?

Largest 4 digit no. = 9999

On dividing 9999 with 15, it leaves quotient of 666 & remainder of 9.

Therefore, 9999 9 = 9990 is the largest multiple of 15 If 9999 is added with 6, it gives the remainder of 6.

Hence, 9996 is the largest 4 digit no. that leaves the remainder of 6.

9. What least number must be added to 3000 to obtain a number exactly divisible by

19?

Sol. On dividing 3000 by 19, we get 17 as remainder.

Number to be added = (19 - 17) = 2.

10. A number being successfully divided by 3, 5 and 8 leaves remainders 1, 4 and 7

Respectively. Find the respective remainders if the order of divisors be reversed.

3 X

5 Y - 1

8 Z - 4

1 - 7

z = (8 x 1 + 7) = 15; y = {5z + 4) = (5 x 15 + 4) = 79; x = (3y + 1) = (3 x 79 + 1) = 238.

Now,

8 238

5 29 - 6

3 5 - 4

1 - 9,

Respective remainders are 6, 4, 2

11. The product of 4 consecutive even numbers is always divisible by?

A.600 B.768 C.864 D.384

Solution:

The product of 4 consecutive 4 numbers is always divisible by 4!.

Since, we have 4 even numbers, we have an additional 2 available with each

number.

It is always divisible by (2^4)4!=16(24)= 384.

12. Find the remainder when 2^99 is divided by 99?

Solution:

power of 2 reminder

1 2

2 4

3 8

4 16

5 32

6 64

7 128 = 29

8 58

9 116 = 17

power of 2 reminder

10 34

11 68

12 136 = 37

13 74

14 148 = 49

15 98

16 196 = 97

17 194 = 95

18 190 = 91

that the remainder when 2^15 is divided by 99 is 98:

2^15 = 99n + 98

I can write this also as

2^15 = 99(n+1) - 1

so the remainder 98" can be considered as a remainder of -1.

But then I know that when I multiply 2^15 times 2^15, the remainder

will be (-1) times (-1), which is 1.

So I know that, (2^15)(2^15) = 2^30

has a remainder 1 when divided by 99.

2^99 = (2^30)(2^30)(2^30)(2^9)

and, since the remainder when 2^30 is divided by 99 is 1, the remainder when 2^99

is divided by 99 is the same as the remainder when 2^9 is divided by 99 - and we

found in our table that this remainder is 17.

NOTE: Suppose we know that 2^n leaves remainder r when divided by 99. Then

we know that

2^n = 99a + r for some integer a

Now consider the next power of 2, which is 2^(n+1). We have

2^(n+1) = 2*(2^n) = 2*(99a + r) = (2*99)a + 2r

The (2*99)a is clearly divisible by 99, so the remainder when 2^(n+1)

is divided by 99 is the remainder when 2r is divided by 99. We have

two possible cases:

(1) if 2r < 99, then the remainder when 2^(n+1) is divided by 99 is 2r

(2) if 2r > 99, then the remainder when 2^(n+1) is divided by 99 is 2r-99

13. A boy writes all the numbers from 100 to 999. The number of zeroes that he uses

is 'a', the number of 5's that he uses is 'b' and the number of 8's he uses is 'c'. What

is the value of b+c-a?

Solution:

We can see by symmetry b=c and hence all we need to calculate b and a

b=280 and a=180

2ba= 380

14. What is the remainder when 91+92+93+....+98 is divided by 6?

Solution:

6 is an even multiple of 3. When any even multiple of 3 is divided by 6, it will leave

a remainder of 0. Or in other words it is perfectly divisible by 6.

On the contrary, when any odd multiple of 3 is divided by 6, it will leave a

remainder of 3. For e.g when 9 an odd multiple of 3 is divided by 6, you will get a

remainder of 3.

9 is an odd multiple of 3. And all powers of 9 are odd multiples of 3.

Therefore, when each of the 8 powers of 9 listed above is divided by 6, each of them

will leave a remainder of 3.

The total value of the remainder =3+3+....+3(8 remainders)=24.

24 is divisible by 6. Hence, it will leave no remainder.

Hence, the final remainder when the expression 91+92+93+....+98 is divided by 6 will

be equal to '0'.

15. The sum of three digit number is subtracted from the number. The resulting

number is always divisible by.

Solution:

Let the three digit number be 439

Sum of digits =16

Difference =43916=423 which is divisible by 9.

16. A and B are playing mathematical puzzles. A asks B "which whole numbers,

greater than one, can divide all the nine three digit numbers111, 222, 333, 444, 555,

666, 777, 888 and 999?" B immediately gave the desired answer. It was

Solution:

Each of the number can be written as a multiple of 111.

The factors of 111 are 3 and 37

Thus the desired answer is 3,37 and 111

17. What is the remainder let after dividing 1!+2!+3!.....+10! by 7?

Solution:

7!+8!+9!.......+100! is completely divisible by 7.

Now, 1!+2!+3!....+6!=873

When 873 is divided by 7 it leaves a remainder = 5.

18. The number 3 divides a with a result of b and a remainder of 2. The number 3 divides b with a result of 2 and a remainder of 1. What is the value of a?

Solution:

Since 3 divides b with a result of 2 and a remainder of 1, b=23+1=7.

Since number 3 divides a with a result of b (which we now know equals 7) and a

remainder of 2, a=b3+2=73+2= 23.

19. It is being given that (232 + 1) is completely divisible by a whole number. Which

of the following numbers is completely divisible by this number?

A. (216 + 1) B. (216 - 1) C. (7 x 223) D. (296 + 1)

Solution:

Let 232 = x. Then, (232 + 1) = (x + 1). Let (x + 1) be completely divisible by the natural number N. Then, (296 + 1) = [(232)3 + 1] = (x3 + 1) = (x + 1)(x2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.

20. How many 3-digit numbers do not have an even digit or a zero?

Solution:

There are 5 digits that are not even or zero: 1,3,5,7 and 9.

Now, lets count all the three-digit numbers that can be formed from these five digits.

The first digit of the number can be filled in 5 ways with any one of the mentioned

digits.

Similarly, the second and third digits of the number can be filled in 5 ways.

Hence, the total number of ways of forming the three-digit number is 125(=555).

21. The LCM of two number is 4800 and their HCF is 160. If one number is 480,

then the second number is

Solution:

HCF LCM = product of two numbers

Hence,4800160=480x

x= 1600

22. N is the smallest number that has 5 factors. How many factors does (N - 1)

have?

Solution:

A number that has 5 factors has to be of the form P4 where 'p' is a prime number.

The smallest such number is 24=16

Therefore, N1=15. The factors of 15 are 1,3,5,15.

So, N - 1 has 4 factors.

23. Which number will completely divide (461 + 462 + 463 + 464) always?

Solution:

The unit value:

For x=odd no:(1,3,5,..), 4 raised to x = 4

For x=even no:(2,4,6,..), 4 raised to x = 6

In here unit values are 4,6,4 and 6

So 4+6+4+6=20, The unit value is 0

So it is divisible by 10

(Or) (461 + 462 + 463 + 464) = 461 x (1 + 4 + 42 + 43) = 461 x 85

= 460 x (4 x 85)

= (460 x 340), which is divisible by 10.

24. (22 + 42 + 62 + ... + 202) = ?

Solution:

(22 + 42 + 62 + ... + 202) = (1 x 2)2 + (2 x 2)2 + (2 x 3)2 + ... + (2 x 10)2

= (22 x 12) + (22 x 22) + (22 x 32) + ... + (22 x 102)

= 22 x [12 + 22 + 32 + ... + 102]

Ref: (12 + 22 + 32 + ... + n2) = 1

n(n + 1)(2n + 1)

6

=

4 x 1

x 10 x 11 x 21

6

= (4 x 5 x 77)

= 1540.

25. Which of the following numbers will completely divide (325 + 326 + 327 + 328)?

A. 11 B. 16 C. 25 D. 30

Solution: (325 + 326 + 327 + 328) = 325 x (1 + 3 + 32 + 33) = 325 x 40

= 324 x 3 x 4 x 10 = (324 x 4 x 30), which is divisible by30.

26. (112 + 122 + 132 + ... + 202) = ?

Solution: (112 + 122 + 132 + ... + 202) = (12 + 22 + 32 + ... + 202) - (12 + 22 + 32 + ... + 102)

Ref: (12 + 22 + 32 + ... + n2) = 1

n(n + 1)(2n + 1)

6

=

20 x 21 x 41 -

10 x 11 x 21

6 6

= (2870 - 385)

= 2485.

27. In a division sum, the divisor is 10 times the quotient and 5 times the

remainder. If the remainder is 46, what is the dividend ?

Solution: Divisor = (5 x 46) = 230

10 x Quotient = 230 = 230

= 23 10

Dividend = (Divisor x Quotient) + Remainder

= (230 x 23) + 46 = 5290 + 46 = 5336.

28. What will be remainder when (6767 + 67) is divided by 68 ?

Solution: (xn + 1) will be divisible by (x + 1) only when n is odd.

(6767 + 1) will be divisible by (67 + 1)

(6767 + 1) + 66, when divided by 68 will give 66 as remainder.

29.

Solution:

30. Which of the following numbers will completely divide (319 + 320 + 321 + 322)?

Solution:319 + 320 + 321 + 322

= 319(1 + 3 + 32 + 33) = 319 40 = 318 3 4 10

= 318 12 10 which is divisible by 12

31. A forester wants to plant 44 apples tree, 66 banana trees and 110 mango trees

in equal rows (in terms of number of trees). Also, he wants to make distinct rows of

tree (i.e. only one type of tree in one row). The number of rows (minimum) that

required is:

Solution: In such case, we first need to find the HCF of 44, 66, 110.

44=2*2*11, 66=2*3*11, 110=2*5*11, HCF= 2*11= 22.

Then, required numbers of rows,={(44/22)+(66/22)+(110/22)}=10.

32. If 146! Is divisible by 5^n, and then find the maximum value of n.

Solution: = [146/5] + [146/5^2] + [146/5^3] = 29+5+1. = 35.

33. The number of numbers from 1 to 200 which are divisible by neither 3 nor 7 is:

Solution: Number of numbers, which are (divisible by 3 + divisible by 7 - divisible by

21), Number of number divisible by 3,= [(198 - 3) /3] +1 = 66

Number of numbers divisible by 7,= [(196 - 7) /7] +1 = 28

Number of number divisible by 21,= [(189 - 21) /21] +1 = 9

Thus, the divisible value = (66 +28 -9) = 85

Thus, number of number which are not divisible by 3 or 7 = 200 -85 = 115

34. A young girl counted in the following way on the fingers of her left hand. She

started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little

finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index

finger 8, thumb 9 and then back to the index finger for 10, middle finger for 11 and

so on. She counted up 1994. She ended on her.

Solution: If the girl counts the way as given in the question, the counting serial for

thumb will be 1, 9, 17, 25 . . . . Hence, number 1992 ( as 1992 is divisible by 8,

common difference of series formed by counting) will also fall on thumb. Hence,

number 1994 will end on her middle finger.

35. Find the remainder when 10 +10^2+10^3+10^4+............. + 10^99is divided by 6.

Solution: The remainder when 10 is divided by 6 is 4.

The remainder when 10^2is divided by 6 is 4.



Thus, remainder is always 4.

So, the required remainder,

= [4 +4 +4 +4 + ........ 99 times] / 6

396 /6. Thus, Required remainder is 0.

36. What is the remainder, when 2256 is divided by 17?

Sol. We can write 17 as 24 + 1 and 2256 as (24)64.

[If f(x) is divided by (x a), the remainder is f(a)] The remainder is ( 1)64 = 1.

37. How many zeroes are at the end of 1000! A. 200 B. 240 C. 248 D. 249

Sol: Since a zero at the end means a factor of ten, we need to determine how many

factors often are contained in 1000! Ten factors as 2*5, and since there will be more

factors of 2, we just need to count the factors of five.

1000/5 = 200, so 200 factors of 5

Furthermore, there are additional factors arising from higher powers of 5.

1000/25 = 40, so 40 additional factors of 5

1000/125 = 8, so 8 additional factors of 5

One factor of 625 also, so one additional factor.

Total: 200 + 40 + 8 + 1 = 249 factors of 5.

Consequently, 1000! ends in 249 zeros.

38. If 100 ! = 100 x 99 x 98 x x 2 x 1, the maximum power of 20 which will divide 100 ! is

A. 21 B. 22 C. 23 D. 24

20 can be written as 2^2 * 5. In 100! we have several even factors. Therefore we

need not check for 2. no of factors of 5 will determine the divisibility. there are 20

multiples of 5. plus 4 factors of 25(100, 75, 50, 25). Each of them giving one extra

multiple of 5. Therefore total of 24 factors of 5.

39. What is the minimum number of square marbles required to tile a floor of

length 5 metres 78 cm and width 3 metres 74 cm?

A. 176 B. 187 C.540 D. 748

The marbles used to tile the floor are square marbles. Therefore, the length of the

marble = width of the marble.

As we have to use whole number of marbles, the side of the square should a factor of

both 5 m 78 cm and 3m 74. And it should be the highest factor of 5 m 78 cm and 3m

74.

5m78cm=578cm and 3m74cm=374cm.

The HCF of 578 and 374 = 34.

Hence, the side of the square is 34.

The number of such square marbles required = 578374/ 3434

=1711= 187 marbles

Documents

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