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Divisibility Rules for CAT Quantitative Aptitude Preparation

Wednesday, February 20th, 2019

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The concept of ‘divide and conquer’, derived from the Latin phrase ‘Divide et impera’, was put into use effectively by everyone from Caesar to Napoleon to The British in India. Even Gaddafi tried using the same but as current events show us – he wasn’t very effective. Dividing rather divisibility rules to be specific can come in really handy at times in solving problems based on Number Systems.

The standard rules which nearly all of us are very comfortable with are the ones for 2ⁿ and 5ⁿ. For these all that one needs to do is look at the last ‘n’ digits of the number. If the last ‘n’ digits of a number are divisible by 2ⁿ or 5ⁿ, then the number is divisible by 2ⁿ or 5ⁿ and vice versa. For details about other numbers, I suggest that you read on.

Funda 1 of Divisibility Rules:

 For checking divisibility by ‘p’, which is of the format of 10ⁿ – 1, sum of blocks of size ‘n’ needs to be checked (Blocks should be considered from the least significant digit ie the right side). If the sum is divisible by p, then the number is divisible by p.
Eg 1.1 Check if a number (N = abcdefgh) is divisible by 9

•   9 is 10¹ – 1
•   Sum of digits is done 1 at a time = a + b + c + d + e + f + g + h = X
•   If X is divisible by 9, N is divisible by 9
•   Also, N is divisible by all factors of 9. Hence the same test works for 3.

Eg 1.2 Check if a number (N = abcdefgh) is divisible by 99

•   99 is 10² – 1
•   Sum of digits is done 2 at a time = ab + cd + ef + gh = X
•   If X is divisible by 99, N is divisible by 99
•   Also, N is divisible by all factors of 99. Hence the same test works for 9, 11 and others.

Eg 1.3 Check if a number (N = abcdefgh) is divisible by 999

•   999 is 10³ – 1
•   Sum of digits is done 3 at a time = ab + cde + fgh = X
•   If X is divisible by 999, N is divisible by 999
•   Also, N is divisible by all factors of 999. Hence the same test works for 27, 37 and others.

Funda 2 of Divisibility Rules:

 For checking divisibility by ‘p’, which is of the format of 10ⁿ + 1, alternating sum of blocks of size ‘n’ needs to be checked (Blocks should be considered from the least significant digit ie the right side). If the alternating sum is divisible by p, then the number is divisible by p.
(Alternating Sum: Sum of a given set of numbers with alternating + and – signs. Since we are using it to just check the divisibility, the order in which + and – signs are used is of no importance.)

Eg 1.1 Check if a number (N = abcdefgh) is divisible by 11

•   11 is 10¹ + 1
•   Alternating sum of digits is done 1 at a time = a  – b + c – d + e – f + g – h = X
•   If X is divisible by 11, N is divisible by 11

Eg 1.2 Check if a number (N = abcdefgh) is divisible by 101

•   101 is 10² + 1
•   Alternating sum of digits is done 2 at a time = ab – cd + ef – gh = X
•   If X is divisible by 101, N is divisible by 101

Eg 1.3 Check if a number (N = abcdefgh) is divisible by 1001

• 1001 is 10³ + 1
•  Sum of digits is done 3 at a time = ab – cde + fgh = X
•  If X is divisible by 1001, N is divisible by 1001
•  Also, N is divisible by all factors of 1001. Hence the same test works for 7, 11, 13 and others.

Funda 3 of Divisibility Rules: 

Osculator / seed number method

For checking divisibility by ‘p’,
Step 1: Figure out an equation such that

If we have this equation, the osculator / seed number for ‘p’ will be . (-m in case of 10m+1 and +m in case of 10m – 1)

Step 2: Remove the last digit and multiply it with the seed number.

Step 3: Add the product with the number that is left after removing the last digit.

Step 4: Repeat Steps 2 and 3 till you get to a number which you can easily check that whether or not it is divisible by p.

Eg: Check whether 131537 is divisible by 19 or not.

•   19*1 = 10*2 – 1 (Seed number is +2)
•  13157-> 13153+7*2 = 13167 -> 1316+7*2 = 1330 -> 133+0*2 = 133
•  133 is divisible by 19
•  131537 is divisible by 19

I hope that these divisibility rules will enable you to divide and conquer few of the Number Systems problems that you encounter during your preparation.

Other posts related to Quantitative Aptitude – Modern Maths

Permutation and Combination – Fundamental Principle of Counting
Permutation and Combination – Distribution of Objects
How to find Rank of a Word in Dictionary (With or Without Repetition)
Set Theory- Maximum and Minimum Values
How to solve questions based on At least n in Set Theory for CAT Exam?
Sequence and Series Problems and Concepts for CAT Exam Preparation
Basic Probability Concepts for CAT Preparation

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