Divisibility questions for CAT covers important CAT divisibility questions in the concept of number systems. This questions will be very useful for the practice of divisibility topic of number systems for CAT.

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**Divisibility Questions for CAT:**

**Question 1:**

How many even integers n, where $100 \leq n \leq 200$ , are divisible neither by seven nor by nine?

A. 40

B. 37

C. 39

D. 38

**Question 2:**

The rightmost non-zero digit of the number $30^{2720}$ is

A. 1

B. 3

C. 7

D. 9

**Question 3:**

Let N = 1421 * 1423 * 1425. What is the remainder when N is divided by 12?

A. 0

B. 9

C. 3

D. 6

**Question 4:**

The integers 34041 and 32506 when divided by a three-digit integer n leave the same remainder. What is n?

A. 289

B. 367

C. 453

D. 307

**Question 5:**

Let N = $55^3 + 17^3 – 72^3$. N is divisible by:

A. both 7 and 13

B. both 3 and 13

C. both 17 and 7

D. both 3 and 17

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**Answers and Questions for Divisibility CAT 2017:**

**Answer 1:**

Between 100 and 200 both included there are 51 even nos. There are 7 even nos which are divisible by 7 and 6 nos which are divisible by 9 and 1 no divisible by both. hence in total 51 – (7+6-1) = 39

There is one more method through which we can find the answer. Since we have to find even numbers, consider the numbers which are divisible by 14, 18 and 126 between 100 and 200. These are 7, 6 and 1 respectively.

**Answer 2:**

Rightmost non-zero digit of $30^{2720}$ is same as rightmost non-zero digit of $3^{272}$.

272 is of the form 4k.

All $3^{4k}$ end in 1.

=> Right most non-zero digit is 1.

**Answer 3:**

The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively.

5*7*9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3

**Answer 4:**

The difference of the numbers = 34041 – 32506 = 1535

The number that divides both these numbers must be a factor of 1535.

307 is the only 3 digit integer that divides 1535.

**Answer 5:**

$55^3 + 17^3 – 72^3$ = $(55-72)k + 17^3$. This is divisible by 17

Remainder when $55^3$ is divided by 3 = 1

Remainder when $17^3$ is divided by 3 = -1

Remainder when $72^3$ is divided by 3 = 0

So, $55^3 + 17^3 – 72^3$ is divisible by 3

So, the answer is d) 3 and 17

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