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 2k
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 5k: If the number formed by the last k digits is divisible by 5k
• By Other Composite Numbers:
In this case we need to check for divisibility by each prime power that divides the given number.
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.
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