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 (= 2⊃2;): If the number formed by the last two digits is divisible by 4

By 8 (=2⊃3;): 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 (= 5⊃2;): If the number formed by the last two digits is divisible by 25

By 125 (= 5⊃3;): 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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