Arbit Numbers

Numbers Theory For XAT 2011 Exam

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NUMBERS PRIME NUMBERS-

Prime number is a natural number greater than unity that has only two factors the number itself and the unity. (e.g. 2,3,5,7,….)

Ø The smallest prime number is 2. Ø There is only one even prime number that is 2. Ø If p1 and p2 are two prime numbers > 2 then p1

2-p22 and p1

2+p22 is composite number.

Ø All prime numbers >3 can be represented in the form of 6n+1 or 6n-1, but reverse of this is not always true, that means all numbers in the form of 6n+1 and 6n-1 may be or may not be a prime number.

RELATIVE PRIMES OR CO-PRIMES- A pair of numbers are said to be relative prime or co-prime to each other if they do not have any common factor other than 1.e.g 15 and 16. List down all the factors of 15 and 16, factors of 15= 1,3,5,15 and that of 16=1,2,4,8,16 so we can see that these two number have only one common factor i.e 1 hence these two are co-prime to each other.

Ø Two consecutive numbers n and n+1 is always co-prime to each other. Ø Squares of two co-prime numbers are always co-prime to each other.

DECIMALS- Decimals can be subdivided in three parts

Ø Terminating decimals- A decimal number that terminate at a point, that means after that point only zero comes. e.g 4.25, 6.5432, 2.3. it has an end or terminating point hence it is called a terminating decimal.

Ø Recurring decimals- It’s a type of decimals in which a digit or a set of digits is repeated continuously. Recurring decimals are written in abridged form, the digits that are repeated being marked by a bar placed over it.e.g 3.666666…..=3.6, or 2.347347347= 2.347. etc.

Ø Non recurring non terminating decimals- Those decimals that does not form any pattern or irrational in pattern are called non recurring non terminating decimals.e.g.√2, √3, √7,π, e, etc.



Ø N.B- So to convert a mixed or impure recurring decimal into fraction, in the numerator write down the entire number formed by none recurring and recurring numbers and subtract from it the part of the decimal that is not recurring. And in the denominator write as many nines as the number of digits having recurring on it and then next to it write as many zeroes as there are digits without recurring in the given decimal.

Ø Hence 0.ab= (ab-a)90, 0.abc= (abc-a)/990, 0.abc= (abc-bc)/900, 0.abcd= (abcd-ab)/9900 SOME IMPORTANT POINTS REGARDING DIVISIBILITY-

Ø Out of a group of n consecutive whole numbers one and only one number is divisible by n.

Ø Difference between any number (of two or more digits) and the number formed by reversing its digits is always divisible by 9.

Ø The product of any n consecutive numbers is always divisible by n!. Ø For any number n (np-n) is always divisible by p where p is a prime number. Ø The square of an odd number when divided by 8 always leaves a remainder 1. Ø For any natural number n5 or n4k+1 is having same unit digit as n has. Where k is a whole

number.

HCF AND LCM

SOME IMPORTANT POINTS FOR HCF AND LCM-

Ø For two numbers a and b their product is equal to the product of their LCM and HCF hence LCM X HCF= ab

Ø If HCF of two numbers is given as ‘h’ then the numbers can be assumed as ‘ha’ and ‘hb’. Ø HCF of a given set of numbers is always a factor of LCM. Ø If ‘h’ is the HCF of two numbers ‘a’ and ‘b’ then ‘h’ is also a factor of n1a+n2b where n1

and n2 are integers. Ø If HCF of two numbers ‘a’ and ‘b’ is ‘h’ then ‘h’ is also HCF of ‘a’ and ‘a+b’ Ø If HCF of two numbers ‘a’ and ‘b’ is ‘h’ then ‘h’ is also HCF of ‘a’ and ‘a-b’ Ø If HCF of two numbers ‘a’ and ‘b’ is ‘h’ then ‘h’ is also HCF of ‘a-b’ and ‘a+b’



QUESTIONS ON HCF AND LCM-

LCM TYPE-1-

When a number divided by a, b or c leaving same remainder r in each case then that number must be in the form of k(LCM of a, b, & c) + r, where k is a natural number with k=1 will give smallest such number greater than the numbers. And if k=0 then we will get the number as remainder.

Ex) Find the smallest number greater than 6 and 8 which when divided by 6 or 8 it leaves a remainder 3 in each case.

Soln) Any number or group of numbers that satisfy the above condition can be given by K(LCM of 6 & 8)+3, since LCM of 6 & 8= 24 so any number in the form of 24k+3 will satisfy above condition, with smallest such number more than the divisors we will when k=1, i.e 27.

Ex) Find the largest 3 digit number which when divided by 5 or 8 it leaves a remainder 4 in each case.

Soln) Any number or group of numbers that satisfy the above condition can be given by K(LCM of 5 & 8)+4, since LCM of 5 & 8= 40 so any number in the form of 40k+4 will satisfy above condition, with smallest such number more than the divisors we will when k=1, i.e 44. But here we need to find largest three digit number that satisfies above condition or in the form of 40k+4. So, 1st we need to find largest 3 digit number that is divisible by 40. When we divide largest 3 digit number 999 by 40 we will get a remainder 39 hence largest 3 digit number that is divisible by 40 or in the form of 40k is 999-39=960. So, largest 3 digit number in the form of 40k+4=964.

LCM TYPE-2-

When a number divided by a, b or c leaving remainders p, q & r respectively such that the difference between divisor and remainder in each case is same i.e (a-p)= (b-q) = (c-r) = t (say) then that number must be in the form of k(LCM of a, b, & c) - t, where k is a natural number with k=1 will give smallest such number greater than the divisors. And if k=0 then we will get the number as remainder.

Ex) Find the smallest 4 digit number that when divided by 3, 5 or 7 gives remainder 1, 3, and 5 respectively.



Soln) Here 3-1=5-3=7-5=2 that means the difference between divisor and remainder is same hence any such number that satisfy above condition must be in the form of k(LCM of 3, 5 &7)-2, Or 105k-2. We need to find smallest 4 digit number in the form of 105k-2. When smallest 4 digit number that is 1000 divided by 105 we will get a remainder 55 that means 1000-55=945 is divisible by 105 but that is largest 3 digit number, and the next multiple of 105 is smallest 4 digit number that is divisible by 105, hence smallest 4 digit number that is divisible by 105 is 945+105= 1050 and required number is 1050-2= 1048.

Here in this case it is easy to visualize that 1050 is divisible by 105 so the required number is 1050-2=1048

LCM TYPE-3-

In the last two cases we have seen some definite pattern now we will see if there is no definite pattern that means neither remainder is same nor the difference between divisor and remainder is same for different divisors. This type of question is solved by trial and error method. We will how to approach such type of question by an example.

Ex)- Find the largest 3 digit number that when divide by 8 or 5 gives a remainder 4 and 3 respectively.

Soln) A number when divided by 8 gives a remainder of 4 hence that number must be in the form of 8n+4 and this number when divided by 5 gives a remainder of 3 hence when subtracted by 3 the result must be divisible by 5 hence 8n+1is divisible by 5, now by putting the values of n =0, 1, 2…. And find out which value of n gives a number that is divisible by 5. We will find n=3 will give 3 X8+1=25 divisible by 5, hence n=3 will give us smallest such number and that is 8n+4= 28. And any number that satisfy above condition must be in the form of k(LCM of 5 & 8) + Smallest number that satisfies the condition. Hence in this case any number that satisfies the condition is in the form of 40k+28. Largest 3 digit number that is divisible by 40 is 960 so largest 3 digit number that satisfies the above condition is 960+28= 988.



HCF TYPE-1-

The largest number with which when divide the numbers a, b, and c the remainders are same then that largest number is given by HCF of (a-b) and (b-c). Here we concern with positive value of the difference. Point to note that HCF of (a-b) and (b-c)= HCF of (a-b) and (a-c) = HCF of (b-c) and (a-c).

Now we will have a brief look why it works. Let common remainder as r and the largest such number as h then when a divided by h it gives a remainder r hence (a-r) is divisible by h similarly (b-r) and (c-r) is divisible by h. We know that if numbers a and b is divisible by another divisor d then (a-b) will also be divisible by d. So working on the same logic take difference of the two in a pair we will get (a-b) and (b-c) both are divisible by h hence largest value of h is the HCF of (a-b) and (b-c) and so on.

HCF TYPE-2-

The largest number with which when the numbers a, b, and c divided giving remainders as p, q, r respectively is given by HCF of three number (a-p), (b-q) and (c-r).

Now before taking an example to understand this 1st we will see why it is HCF of the (a-p), (b-q) and (c-r). Let largest such number is h, so when a is divided by h it gives a remainder p hence (a-p) must be divisible by h. Similarly (b-q) and (c-r) also divisible by h, hence largest value of h has to be HCF of (a-p), (b-q) and (c-r).

Ex) Find the largest number with which when 182, 228 and 275 are divided remainders are 2, 3 and 5 respectively.

Soln) Let such a number is h hence 182-2=180 is divisible by h and similarly 225 and 270 is divisible by h. So, largest value of h is given by HCF of 180, 225 and 270 and that is 45.

LCM & HCF OF FRACTIONS-

LCM- LCM of two or more fractions is given by

LCM of numerators of all the fractions

HCF of denominators of all the fractions



HCF- HCF of two or more fractions is given by

HCF of numerators of all the fractions

LCM of denominators of all the fractions

PROPERTIES OF FACTORS OF A NUMBER

NUMBER OF FACTORS OF A NUMBER-

To find number of factors of a composite number write down the number in the form of n=apbqcr…. where a, b and c are prime numbers , and p, q and r are natural number then number of factors of n is given by (p+1)(q+1)(r+1)….. and it includes 1 and number itself.

It is important to note that if numbers of factors are odd then the number is a perfect square and if number of factors is even then number is not a perfect square. This is because if number is perfect square then p, q, and r is even and hence (p+1) (q+1) and (r+1) are odd and so product of these numbers is also a odd number.

NUMBER OF WAYS TO EXPRESS A NUMBER AS A PRODUCT OF TWO FACTORS

To find out the number of ways to express a number as products of two factors 1st convert the number in the form of n=apbqcr….. then total number of factors is (p+1)(q+1)(r+1)… and number of ways to express the number as a product of two numbers are ½(p+1)(q+1)(r+1).

SUM OF ALL THE FACTORS OF A NUMBER-

If we need to find out the sum of all the factors of a number then 1st write down the number in the form of n=apbqcr….. as we all know here a, b, c, are prime numbers, then sum of all the factors of the number n= {(ap+1-1)/(a-1) X (bq+1-1)/(b-1) X (cr+1-1)/(c-1)…..}



NUMBER OF WAYS OF WRITING A NUMBER AS PRODUCT OF 2 COPRIMES-

If we need to find the number of ways of writing a number as a product of two co-primes then 1st write down the number in the form of n=apbqcr….. where a, b c .. are prime numbers if in this expression we have k numbers of prime numbers then required number of ways= 2k-1.

NUMBER OF CO-PRIMES TO A NUMBER THAT ARE LESS THAN THE NUMBER-

The number of co-primes to number n and also less than n are given by n(1-1/a)(1-1/b)(1-1/c)…… where a, b, c… are the prime numbers when number n is written in the form of n=apbqcr…..

TO FIND LAST DIGIT

In general cases where we need to find out the last digit of certain number of power to a number. It is property of numbers that unit digit of powers repeat after certain period we just need to find out that period once we know the period we can find out unit digit of any power. Here we will see this property for some numbers.

Here is the table for unit digit of an.

n(→) 1 2 3 4 Cyclicity

a(↓)

0 0 0 0 0 1

1 1 1 1 1 1

2 2 4 8 6 4

3 3 9 7 1 4

4 4 6 4 6 2

5 5 5 5 5 1

6 6 6 6 6 1

7 7 9 3 1 4

8 8 4 2 6 4

9 9 1 9 1 2



The fifth power of any number has the same units place digit as the number itself.

Always remember one formula in cyclic method if you get 1 then that is the terminating point and after that cycle will start again as we have seen in the case of 7 since 74 has unit digit is 1 then it has to repeat from 5th power and cycle has a period of 4.

REMAINDER OF A DIVISION

To find out the remainder of any division we have basically 2 methods, but it can be extended to the binomial theorem as well. These two important methods are as follows.

1) PATTERN METHOD 2) REMAINDER THEOREM.

PATTERN METHOD- This method is also called cyclic method and implication of this method is same as what we did in last section to find out the unit digit.

REMAINDER THEOREM-

We can use remainder theorem to find out the remainder in the problems we have discussed above. So 1st let us see what is Remainder Theorem.

Remainder theorem states that when a function f(x) a polynomial function divided by (x-a) where a is a constant, will give us a remainder f(a).

Let us make it more clearly by examples-

Ex) Find the remainder of (2x2-3x-4)/(x-3).

Soln) Here given function is f(x)= 2x2-3x-4, and it is divided by (x-3) hence a=3, so remainder of this division is f(3)= 2 X 32-3 X 3-4= 18-9-4=5.

N.B- If remainder comes is negative then to get actual remainder just add the remainder to the divisor.



PROPERTIES OF FACTORIAL!

Factorial of a natural number is defined as the products of all the integers from 1 to n. It is represented as ! or /_. So factorial 4 can be written as 4! = 1 X 2 X 3 X 4 = 24.

So n!= 1 X 2 X 3 X…… X n, and 0! Is defined as 1 and we will see the proof of this in the chapter of permutation and combination.

Most of the questions in the numbers related to factorial are of the type to find out the largest power of a number in a factorial n (n!).

Highest power of a PRIME NUMBER p that divides n! exactly that means without leaving any remainder is given by [n/p]+ [n/p2] + [n/p3]……. Where, [x] represents greatest integer function, that means greatest integer less than or equal to x. We will see that with some examples.

For a composite number, convert the number as product of primes and find the highest power of the largest prime factor that will be the highest power of the composite number in any given factorial.

NUMBER OF ZEROS IN A NUMBER-

Number of zeros in a number is same as the highest power of 10 in that number, and we know that 10=2 X 5 then we needs to find the highest power of 2 & 5, and smaller in the two is equal to highest powers of 10.



SOME OTHER PROPERTIES OF NUMBERS

SQUARES- There is a definite relationship between the unit digit when square of a number is considered, we will see that one by one.\

Ø If unit digit of a perfect square is 1 then ten’s digit has to be Even (e.g 81, 121, 441 etc) Ø If unit digit of a number is 2 then it cant be a perfect square. Ø If unit digit of a number is 3 then it cant be a perfect square. Ø If unit digit of a perfect square is 4 then ten’s digit has to be Even (e.g 64, 144 etc) Ø If unit digit of a perfect square is 5 then ten’s digit has to be 2 (e.g 25, 225, 625 etc) Ø If unit digit of a perfect square is 6 then ten’s digit has to be Odd and multiple of 3. Ø If unit digit of a number is 7 & 8 then it cant be a perfect square Ø If unit digit of a perfect square is 9 then ten’s digit has to be Even (e.g 49, 169 etc) Ø If unit digit of a perfect square is 0 then ten’s digit has to be 0 (e.g 100, 400, 900 etc)

Hence if a number end with 2, 3, 7 or 8 then it cant be perfect square.

If a perfect square is written in the form of n=apbqcr… then all the powers p, q, r… has to be even.

TO FIND SQUARE OF A NUMBER- It is our sincere advice for all students to remember the squares of 1st 25 natural number, and if you remembered then get ready for shortcut to find the squares of other digits-

Squares for numbers from 26 to 50- Let you need to find the square of a two digit

number ab from 25 to 50. Find 50-ab= a two digit number cd, where cd has to be in between 1-25 (and square of that you know). Find the square of cd and write down last two digits only and carry over the rest digit let that carry over is x then for rest digits of the result find out ab+x-25. We will see that with the help of examples.



Ex) Find the square of 37.

Soln) here given two digit number ab=37, find out 50-37=13, so cd=13, now find the square of 13, 132= 169, so write down 69 as last two digit and carry over 1, so here x=1, for rest digits calculate ab+x-25= 37+1-25=13, so square of 37= 1369.


number ab from 51 to 75. Find ab-50= a two digit number cd, where cd has to be in between 1-25 (and square of that you know). Find the square of cd and write down last two digits only and carry over the rest digit let that carry over is x then for rest 1st two digits of the result find out ab+x-25. We will see that with the help of examples.


number ab from 75 to 100. Find 100-ab= a two digit number cd, where cd has to be in between 1-25 (and square of that you know). Find the square of cd and write down last two digits only and carry over the rest digit let that carry over is x then for rest 1st two digits of the result find out 2ab+x-100. We will see that with the help of examples.

SOME IMPORTANT ALGEBRIC FORMULAE

a3 ± b3 = (a ± b)(a2 + ab + b2). Hence, a3 ± b3 is divisible by (a ± b) and (a2 ± ab + b2). an - bn = (a – b)(an-1 + an-2b+ an-3b2 + ... + bn-1)[for all n]. Hence, an - bn is divisible by a - b for all n. an - bn = (a + b)(an-1 – an-2b + an-3b2 ... – bn-1)[n-even] Hence, an - bn is divisible by a + b for even n. an + bn = (a + b)(an-1 – an-2b + an-3b2 + ... + bn-1)[n-odd] Hence, an + bn is divisible by a + b for odd n. a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - ac - bc) Hence, if a + b + c = 0, a3 + b3 + c3 = 3abc



SOME IMPORTANT PROPERTIES ►Wilson’s Theorem If p is a prime number then (p-1)! +1 is divisible by p ►Fermet’s Little Theorem If P is a prime number then (ap – a) is divisible by p ► Any single digit number written (P-1) times is divisible by P, where P is a prime number >5. Examples: 222222 is divisible by 7 77777….. 10 times is divisible by 11

LAST TWO DIGITS Last two digit of (ab)n = last two digit of (bn) + (nabn-1) It is very useful formula but it cannot be used in certain cases. Consider some cases- Case (1)- If b=1, then last two digit will be Last two digit of (na+01) Case (2) If b = 3, 7 or 9, then 1st convert it in the form such that unit digit is 1. Ex- Find the unit digit of (37)2000 Ans- Since we know that 74 has unit digit 1, hence find last two digit of (37)4. Now we don’t need to find the exact value of (37)4, Just find last two digit of (37)2 it is 69, then find last two digit of (69)2 it is 61. Hence last two digit of (37)4 is 61 so we can write down (37)2000 = (61)500. Now we can find the last two digit of (61)500. = Last two digit of (6 x 500) + (01) = 01 Case (3) If b=5, then last two digit = Last two digit of (25na) + (25)



XAT (Numbers)

1. Let X be a four-digit positive integer such that the unit digit of X is prime and the product of all digits of X is also prime. How many such integers are possible? [XAT 2010]

(1) 4 (2) 8 (3) 12 (4) 24 (5) None of the above

2. a, b, c, d and e are integers such that 1 ≤ a < b < c < d < e. If a, b, c, d and e are geometric progression and lcm(m, n) is the least common multiple of m and n, then the maximum value of

�

�� +

�

�� +

�

�� +

�

�� [XAT 2010]

(1) 1 (2) 15/16 (3) 79/81 (4)7/8 (5) None of these

3. A chocolate dealer has to send chocolates of three brands to a shopkeeper. All the brands are packed in

boxes of same size. The number of boxes to be sent is 96 of brand A, 240 of brand B and 336 of brand C. These boxes are to be packed in cartons of same size containing equal number of boxes. Each carton should contain boxes of same brand of chocolates. What could be the minimum number of cartons that the dealer has to send? [XAT 2010] (1) 20 (2) 14 (3) 42 (4) 38 (5) 16



Instructions: Consider the information given below for questions 4 and 5. In the diagram below, the seven letters correspond to seven unique digits chosen from 0 to 9. The relationship among the digits is such that: [XAT 2009] P.Q.R = X.Y.Z = Q.A.Y

P X

Q A Y

R Z

4. The value of A is: A. 0 B. 2 C. 3 D. 6 E. None of the above 5. The sum of digits which are not used is: A. 8 B. 10 C. 14 D. 15 E. None of the above

6. Company BELIANCE hosted a party for 8 members of Company AXIAL. In the party no member of

AXIAL had interacted with more than three members of BELIANCE. Out of all the members of BELIANCE, three members – each interacted with four members of AXIAL and the remaining members – each interacted with two members of AXIAL. The greatest possible number of members of company BELIANCE in the party is [XAT 2009]

A. 9 B. 10 C. 11 D. 12 E. None of the above. 7. Let X be a four-digit number with exactly three consecutive digits being same and is a multiple of 9. How many such X's are possible? [XAT 2009]

A. 12 B. 16 C. 19 D. 21 E. None of the above.

8. Four digits of the number 29138576 are omitted so that the result is as large as possible. The largest omitted digit is [XAT 2008]

(A) 9 (B) 8 (C) 7 (D) 6 (E) 5



9. If [X] denotes the greatest integer less than or equal to X, then ��

� + ��

� �

�� + ��

� �

�� + …..+ ��

� ��

�� [XAT 2008]

(A) 33 (B) 34 (C) 66 (D) 67 (E)98

10. In the figure, number in any cell is obtained by adding two numbers in the cells directly below it. For example, 9 in the second row is obtained by adding the two numbers 4 and 5 directly below it. The value of X – Y is. [XAT 2008]

(A) 2 (B) 3 (C) 4 (D) 5 (E) 6



Questions 11 - 12: A, B, C, D, E and F are six positive integers such that B + C + D + E = 4A C + F = 3A C + D + E = 2F F = 2D E + F = 2C + 1 If A is a prime number between 12 and 20, then 11. The value of C is [XAT 2008]

(A) 23 (B) 21 (C) 19 (D) 17 (E) 13 12. The value of F is

[XAT 2008]

(A) 14 (B) 16 (C) 20 (D) 24 (E)28 13. Which of the following must be true? [XAT 2008]

(A) D is the lowest integer and D = 14 (B) C is the greatest integer and C = 23 (C) B is the lowest integer and B = 12 (D) F is the greatest integer and F = 24 (E) A is the lowest integer and A = 13



14. If x > 0, then minimum value of {(x+1/x)6 – (x6 + 1/x6) – 2}/ {(x+1/x)3 + (x3 + 1/x3)} [XAT 2007]

(A) 6 (B) 3 (C) 2 (D) 1 (E) None of these Question 15 to 17: Substitute different digits (0, 1, 2,….9) for different letters in the problem below, so that the

corresponding addition is correct and it results in the maximum possible value of MONEY. P A Y

M E

R E A L

M O N E Y

15. The letter ‘Y’ should be [XAT 2007]

A. 0 B. 2 C. 3 D. 7 E. None of the above

16. There are nine letters and ten digits. The digit that remains unutilized is: [XAT 2007]

A. 4 B. 3 C. 2 D. 1 E. None of the above

17. The resulting value of ‘MONEY’ is [XAT 2007]

A. 10364 B. 10563 C. 10978 D. 19627 E. None of the above 18. If a = 24, b = 26, c = 28, then the value of a2+ b2 + c2 - ab - be - ca will be (A) 0 (B) 4 (C) 8 (D) 12 [XAT 2006]

19. . A number n said to be perfect if the sum of all its divisors (excluding n itself) is equal to n. An example of perfect number is: (A) 9 (B) 15 (C) 21 (D) 6 [XAT 2006]



20. If x = 6 and y = 3, then the value of [x + y] x/y is : (A) 30 (B) 36 (C) 81 (D) 18 [XAT 2006]

21. A set of balls is numbered and the balls are arranged from 1

to 50. Ball number 3 is drawn first and then every 5th ball thereafter is drawn. What will be the number of the last ball

drawn? (A) 48 (B) 46 (C) 47 (D) 50 [XAT 2006]

22. If x4/x9 = 5-5 then x is equal to x

(A) 5 (B) 4 (C) 9 (D) None of these [XAT 2006]

23. How many numbers between 1 to 1000 (both excluded) are both squares and cubes? [XAT 2005]

(A) none (B) 1 (C) 2 (D) 3

24. If x > 8 and y > – 4, then which one of the following is always true? (A) xy < 0 (B) x2 < – y (C) – x < 2y (D) x > y [XAT 2005]

25. For n = 1, 2, .... let Tn = 13 + 23 + ... + n3, which one of the following statements is correct?

(A) There is no value of n for which Tn is a positive power of 2. (B) There is exactly one value of n for which Tn is a positive power of 2. (C) There are exactly two values of n for which Tn is a positive power of 2. (D) There are more than two values of n for which Tn is a positive power of 2.

[XAT 2005]



SOLUTIONS

1. Unit digit of x is prime hence it can be 2, 3, 5, or 7. As the product of all digits of x is prime hence rest all digits must be 1, then possible values of x 1112, 1113, 1115 or 1117. Option (1) 2. To maximize the value of the given expression, denominator should be minimized. Therefore possible values are as follows. a = 1 b = 2 C = 4 d = 8 e = 16 = ½ +1/4+1/8+1/16 = 15/16 So 15/16 is the maximum value. Option (2) 3. Given that number of boxes of brand A, B and C are 96, 240 and 336 respectively. Each carton should contain boxes of same brand and also an equal number of boxes. This can happen only when each carton contains a number of boxes that is a factor of number of boxes of each brand i.e. a factor of 96, 240 and336. Maximum number of boxes in one carton = HCF of 96, 240, 336 = 48. So, numbers of cartons containing boxes of brand A = 96/48 = 2 Numbers of cartons containing boxes of brand B = 240/48 = 5 Numbers of cartons containing boxes of brand C = 336/48 = 7 So, total numbers of cartons = 2 + 5 + 7 = 14 Option (2) 4. Given that all Seven letters correspond to seven unique digits chosen from 0 to 9 and

p.q.r = x.y.z = q.a.y

hence none of them can be 0 as then whole product will become 0, so possible digits must be from 1 to 9.



Now consider 5 it has no other multiple between 1 to 9 hence it is also eliminated.

On the same logic 7 is also eliminated.

Now possible digits are {1, 2, 3, 4, 6, 8, 9}

Consider p.q.r = x.y.z and all six digits are distinct numbers.

So p.q.r = x.y.z = q.a.y = LCM (1, 2, 3, 4, 6, 8, 9) = 22 x 32 = 72 (product must be the LCM of the numbers)

72 = 1 x 8 x 9 = 6 x 4 x 3

x, y, z, p, q, r ∈ {1, 3, 4, 6, 8, 9}, hence only left digit is 2, so a=2.

Option (B).

5. Here 0, 5 and 7 were eliminated hence sum of the digits eliminated = 0+5+7 = 12

Option (E)

6. Given that there are 8 members in AXIAL, and each of them interacts with a maximum of 3 members from BELIANCE.

Maximum number of possible interactions is 8 × 3 = 24.

Assume that there are n members in BELIANCE. Since 3 members of BELIANCE each interact with 4 members of AXIAL hence 12 interactions occur in this process.

Remaining number of interactions are 24 – 12 =

Number of interactions that could occur among the remaining members.= 12

The remaining members in BELIANCE is (n – 3) and each of them interact with 2 members of AXIAL, therefore the number of interactions will be (n – 3) × 2 = 12, that is when n = 9

Option (A)



7. Given that the number should divisible by 9 hence sum of digits should be divisible by 9

It is also given that it is a four digit number.

∴ The sum of 4 digits could be 9, 18, 27 or 36. But it is given that exactly 3 consecutive digits are same and 4th is different, hence sum of digits cannot be 36.

Now we have III cases

Case I → if sum of digits is 9.

Then Possible numbers are: 1116, 6111, 2223, 3222, 3330 and 9000.= total 6 numbers.

Case II →If sum of digits is 18.

Possible numbers are: 3339, 9333, 4446, 6444, 5553, 3555 and 6660. = 7 numbers.

Case III → sum of digits is 27.

Possible numbers are: 6669, 9666, 7776, 6777, 8883, 3888 and 9990. = 7 numbers.

Total possible numbers are 6+7=7 20

Option (E).

8. Given number is 29138576. We have to omit any four digits in such a manner that the remaining number we get is the largest possible number. If we omit 2, 1, 3 and 5 then the result will be 9876 and it is the largest possible. Hence The largest omitted digit will be 5. Option E. 9. Since we know that [x] = greatest integer less than or equal to x. And tn = [1/3 + n/99] where 0≤n ≤98 Consider [1/3] = [0.3333] =0 [1/3 + 1/99] =0 Hence till n = 65 we will have [x] =0 as when n=65 then [1/3 +65/99] =0 and [1/3+66/99] =1



Hence [x] =1 when 65 < n < 99

Hence required summation = {0+0+…(66terms)} + {1+1+…(33 terms)} =33

Option A.

10. We have two variable here and if we know the numbers at the bottom most row then we can calculate the numbers for any row.

Continue filling the boxes from the bottom, we get the above one.

Proceeding upward we get the above one.

Then, Y+29 = 16+ X+ 9

X – Y = 4

Option C.



11.

Here we are provided with5 equations and 6 unknown and extra information about A

hence we will eliminate all variables and try to keep equation in terms of A

From equation (iii) and (iv) eliminate F we will get

C + E = 3D …………….(vi)

From equation (iv) and (v) we get, again we eliminate F

E = 2C- 2D+ 1

3C+1= 5D ⋯(vii)

But rom equation (iv), we get, 3C+1 = 5F/2

F = (6C+2)/5

Put this value in (ii) We get,

11C+2 = 15A

Or c = (15A-2)/11

It is given that A is a prime number between 12 and 20.

Hence possible values of A are 13,17 or 19.

Since C is a positive integer hence 15A must be divisible by 11

When A = 13 then 15A-2 = 195-2 =193 not divisible by 11 hence ruled out.

Given equations are

B + C + D + E = 4A … (i)

C + F = 3A … (ii)

C + D + E = 2F … (iii)

F = 2D … (iv)

E + F = 2C + 1 … (v)



When A = 17, then 15A-2 = 255-2 = 253 it is divisible by 11.

When A =19 then 15A -2 = 285-2= 283, not divisible by 11.

Hence A = 17.

Then the value of C is 23, Option A.

Ans. 12. Refer to the previous question

Since A =17 hence we will get F =28

Option E

Ans. 13. After finding out all the values we will find out that B is the lowest integer and B = 12

Option C

14. we need to find the minimum value of [(x+1/x)6 – (x6 +1/x6) – 2] / [(x+1/x)3 +(x3 +1/x3)] Consider the numerator [(x+1/x)6 – (x6 +1/x6) – 2] = [(x+1/x)6 –(x3 +1/x3)] = [(x+1/x)3 +(x3 +1/x3)] [(x+1/x)3 -(x3 +1/x3)] And denominator is [(x+1/x)3 +(x3 +1/x3)] hence the result of division = [(x+1/x)3 -(x3 +1/x3)] =3(x+1/x) We know that if x>0 then minimum value of (x+1/x) will be 2, Hence the minimum value of that division = 2x3 =6 Option (A)

15.Unit place letter is =Y.so, Sum of Y,L,E must be >10. That means ,E+L=10

Again,A+M+A+ Carry=E=>2A=E+8[ carry amd M=1][ As M is the left most = carry]

N=P+E+1

If E=2 then, MONEY=19627.



OPTION: (D)

16. From the solution of last question

Unutilized remain 4.

OPTION: (A)

17. From the solution of last question

MONEY=19627.

OPTION: (D)

18. (a2+ b2+ c2- ab- bc - ca) = a(a-b) + b(b-c)+ c(c-a)

Now putting the value of a, b and c we get 12

Option : (D)

19. Based on observation and eliminating the options we will get that number as 6. Divisor of 6 are , 1, 2, 3

There sum gives us 6. Option : (D)

20. [x+y]x/y = [6+3]2 = 81

Option : (D)



21. As per question ,

3 + 5th ball(i.e 8 no. Ball) + 5th ball(i.e 13 no. Ball) +..........+48 no. Ball. Hence, last ball no. Drawn be 48. Option (A)

22. As equation can be written as ��

= ��

, x= 5

Option : (A)

23. The number is perfect square and cube both hence power of the corresponding number must be 6, these numbers are 26 = 64, 36 = 729 Option (C)

24. Here x = 9, 10, 11 .... � y = – 3, – 2, – 1, 0, 1, 2, 3, ....

Option (C)

25. Solve it by yourself!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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