# COGNIZANT Divisibility QUESTIONS WITH SOLUTIONS

1. The number of prime factors of (3 x 5)12 (2 x 7)10 (10)25 is:

A): 47
B): 60
C): 72
D): None of these
Correct Answer : D
Explanation: The equation can be facorize as 3*5*3*2*2*2*7*2*5*2*5*5*5 or 2^5*3^2*5^5*7^1 total no of prime factor =(5+1)*(5+1)*(2+1)*(1+1)=216

1. What least value must be assigned to * so that the number 63576*2 is divisible by 8?

A): 1
B): 2
C): 3
D): 4
Correct Answer :C
Explanation: The test for divisibility by 8 is that the last 3 digits of the number in question have to be divisible by 8.
So, 6*2 has to be divisibile by 8.
I know 512 is divisible by 8.
Also 592 is divisible by 8.
So, 632 is divisible by 8.
So * is 3.

1. The smallest number, which is a perfect square and contains 7936 as a factor is:

A): 251664
B): 231564
C): 246016
D): 346016
Correct Answer :C
Explanation:
7936 => 2^2 * 2^2 * 2^2 * 2^2 * 31^1
To make it as a perfect square, we have to multiply 7936 with 31…
Hence the reqd no. is 7936*31 = 246016

1. In a division problem, the divisor is twenty times the quotient and five times the remainder. If remainder is 16, the number will be:

A): 3360
B): 336
C): 1616
D): 20516
Correct Answer :B
Explanation:
Divisor = (5 x 46) = 230

 10 x Quotient = 230     = 230 23 10

Dividend = (Divisor x Quotient) + Remainder
= (230 x 23) + 46
= 5290 + 46
= 5336.

1. If a number is exactly divisible by 85, then what will be the remainder when the same number is divided by 17?

A): 3
B): 1
C): 4
D): 0
Correct Answer : D
Explanation: number=divisor*quotient+remainder
so 17*5+0;
remainder is 0;
divisor is 17;
quotient is 5;

1. The least perfect square number which is exactly divisible by 3, 4, 7, 10 and 12 is:

A): 8100
B): 17600
C): 44100
D): None of these
Correct Answer : C

1. (xn+yn) is divisible by (x-y):

A): for all values of n
B): only for even values of n
C): only for odd values of n
D): for no values of n
Correct Answer : D
Explanation: for no values of n

1. P is an integer. P is greater than 883.If P -7 is a multiple of 11, then the largest number that will always divide (P+4)(P+15) is

A)242
B)343
C)321
D)none
Answer:A
Explanation: p-7= 11*a (as it is multiple of 11)
p=11*(a+7)
so (p+4)(p+15)= (11a+7+4)(11a+7+15);
= (11a+11)(11a+22);
=11*11(a+1)(a+2);
=121*2
=242

1. The greatest number that will divide 63, 138 and 228 so as to leave the same remainder in each case:

A): 15
B): 20
C): 35
D): 40
Correct Answer :A
Explanation: The greatest number = H.C.F of (138-63), (228-138), (228-63)
H.C.F of 75, 90, 165 = 15.
15 is the greatest number.

1. Find the largest number, smaller than the smallest four-digit number, which when divided by 4,5,6and 7 leaves a remainder 2 in each case:

A): 422
B): 842
C): 12723
D): None of these
Correct Answer : B
Explanation: Take LCM of 4,5,6,7. It is 420
BUt the no must leave remainder 2 in each case, so the no is of the form: 420k + 2.
The smallest 4-digit no is 1000. So keeping k=0,1,2,3….
We get that the largest no smaller than the smallest 4 -digit no is 842

1. What is the highest power of 5 that divides 90 x 80 x 70 x 60 x 50 x 40 x 30 x 20 x 10?

A): 10
B): 12
C): 14
D): None of these
Correct Answer : A
Explanation: Take LCM of Each Number
90/5=5*2*3*3——————>here we will get one 5
80/5=5*2*2*2*2—————>here we will get one 5
70/5=5*2*7————–___—->here we will get one 5
60/5=5*2*2*3——————>here we will get one 5
50/5=5*5*2___——————>here we will get Two 5^2
40/5=5*2*2*2——————>here we will get one 5
30/5=5*2*3———————>here we will get one 5
20/5=5*2*2———————>here we will get one 5
10/5=5*2————————>here we will get one 5
Here we will get one 5 in each number instead of 50(5*5*2)
So answer is 5^10

1. If a and b are natural numbers and a-b is divisible by 3, then a3-b3 is divisible by:

A): 3 but not by 9
B): 9
C): 6
D): 27
Correct Answer : B
Explanation: If a − b is divisible by 3, then a − b = 3k, for some integer k
(a − b)² = (3k)²
a² − 2ab + b² = 9k²
a³ − b³ = (a−b) (a² + ab + b²)
= (a−b) (a² − 2ab + b² + 3ab)
= 3k (9k + 3ab)
= 3k * 3 (3k + ab)
= 9 k(3k+ab)
Since k(3k+ab) is an integer, then 9k(3k+ab) is divisible by 9

1. What is the greatest positive power of 5 that divides 30! exactly?

A): 5
B): 6
C): 7
D): 8
Correct Answer : C
Explanation:  The question is, how many powers of 5 are in the factors of 30! (that’s 30factorial, for those above)…
Only the numbers 5, 10, 15, 20, 25, and 30 have divisors of 5. And 25 is divisible by 5^2.
So the answer is 5*5*5*5*(5^2)*5 = 5^7.

1. What is the smallest four-digit number which when divided by 6, leaves a remainder of 5 and when divided by 5 leaves a remainder of 3?

A): 1043
B): 1073
C): 1103
D): None of these
Correct Answer : D
Explanation: remainder when  m is divided by 5  = 2
Smallest m is 2.
Hence, N = 1001 + 6 * 2 = 1013.

1. P is an integer. P>883. If P-7 is a multiple of 11, then the largest number that will always divide (P+4) (P+15) is:

A): 11
B): 121
C): 242
D): None of these
Correct Answer : C
Explanation: Given P is an integer>883.
P-7 is a multiple of 11=>there exist a positive integer a such that
P-7=11 a=>P=11 a+7
(P+4)(P+15)=(11 a+7+4)(11 a+7+15)
=(11 a+11)(11 a+22)
=121(a+1)(a+2)
As a is a positive integer therefore (a+1)(a+2) is divisible by 2.Hence (P+4)(P+15) is divisible by 121*2=242

1. Let C be a positive integer such that C + 7 is divisible by 5. The smallest positive integer n (>2) such that C + n2 is divisible by 5 is:

A): 4
B): 5
C): 3
D): Does not exist
Correct Answer : D
Explanation: c + n^2 is divisible by 5 if and only if c and n^2 are both divisible by 5.
But, if c is divisible by 5 then c + 5 will not be divisible by 5.
So, option(d ) is correct.

1. Four bells begin to toll together and then each one at intervals of 6 s, 7 s, 8 s and 9 s respectively.

The number of times they will toll together in the next 2 hr is:

A): 14 times
B): 15 times
C): 13 times
D): 11 times
Correct Answer : A
Explanation: first we to find the L.C.M. of 6, 7, 8 and 9.
Prime factorization of 6 = 2*3
Prime factorization of 7 = 7
Prime factorization of 8 = 2*2*2
Prime factorization of 9 = 3*3
L.C.M. = 2*2*2*3*3*7
= 504
The L.C.M. of 6 seconds, 7 seconds, 8 seconds and 9 seconds is 504
seconds.
Now, 1 hour = 3600 seconds
So, 2 hours = 3600*2 = 7200 seconds
The number of times the four bells will toll together in the next 2 hour
= 7200/504
= 14.28 or 14 times
They will toll together 14 times in the next 2 hours

1. On dividing a number by 999,the quotient is 366 and the remainder is 103.The number is:

A): 364724
B): 365387
C): 365737
D): 366757
Correct Answer : C
Explanation: Number (Dividend) = Divisor * quotient + remainder.
Number = 999 * 377 + 105 = 3767

1. If 522x is a three digit number with as a  digit x . If the number is divisible by 6, What is the value of the digit x is?

A)1
B)2
C)3
D)4
e)6
Answer:e
Explanation: If a number is Divisiable by 6 , it must be divisible by both 2 and 3
In 522x, to this number be divisible by 2, the value of x must be even. So it       can be 2,4 or 6 from given options
552x is divisible by 3, If sum of its digits is a multiple of 3.
5+5+2+x =12+x   ,
If put x =2 , 12+2=14 not a multiple of 3
If put x =4 , 12+6=18  is a multiple of 3
If put x =6 , 12+2=14 not a multiple of 3
The value of x is 6.
Number System Sample Question with Solutions

1. P is an integer. P is greater than 883.If P -7 is a multiple of 11, then the largest number that will always divide (P+4)(P+15) is

A)242
B)340
C)245
D)178
Answer:A
Explanation: p-7= 11*a (as it is multiple of 11)
p=11*(a+7)
so (p+4)(p+15)= (11a+7+4)(11a+7+15);
= (11a+11)(11a+22);
=11*11(a+1)(a+2);
=121*2
=242