# CAT Quantitative Aptitude Questions - Tests of Divisibility

In the CAT exam, concept of Divisibility holds great importance for the aspirants to gain high percentile in the exam. Read on some examples to gain understanding of the topic in detail and prepare for the CAT exam!

CAT Quantitative Aptitude has questions on Test of Divisibility concepts. A natural number can be checked for divisibility by other natural numbers quickly in special cases as follows: **• By Primes: **

Tests for divisibility by the following primes use the digits of the given number. The number is divisible

By 2: If the last digit is even (i.e. divisible by 2)

By 3: If the sum of digits is divisible by 3

By 5: If the last digit is either 0 or 5 (i.e. divisible by 5)

By 11: If the difference between the sum of the digits in even places and the sum of the digits in odd places is divisible by 11

**• By Powers of Primes:**

By 4 (= 2²): If the number formed by the last two digits is divisible by 4

By 8 (=2³): If the number formed by the last three digits is divisible by 8

By 2
* ^{k}*: If the number formed by the last k digits is divisible by 2

^{k}By 9: If the sum of the digits is divisible by 9

By 25 (= 5²): If the number formed by the last two digits is divisible by 25

By 125 (= 5³): If the number formed by the last three digits is divisible by 125

By 5

*: If the number formed by the last k digits is divisible by 5*

^{k}

^{k}

**• By Other Composite Numbers:**

In this case we need to check for divisibility by each prime power that divides the given number.

For example:

By 6: Check for divisibility by 2 and by 3

By 12: Check for divisibility by 4 and by 3

By 22: Check for divisibility by 2 and by 11 etc.*Examples:** *(1) 475 is divisible by 25 but not by 125.

(2) 465 is divisible by 5 but not by a higher power of 5.

As 4 + 6 + 5 = 15, it is divisible by 3 though.

(3) To check for divisibility of 8976730213 by 11, we first form the two sums: 8 + 7 + 7 + 0 + 1 (= 23) and 9 + 6 + 3 + 2 + 3 (= 23)

then find the difference between the two 23 -23 = 0.

As 0 is divisible by 11, so is the number 8976730213.

(8976730213 = 11 × 816066383).

Now you can apply the concept to solve the toughest CAT Quantitative Aptitude Questions.

**Good Luck for CAT 2018!!**

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