# CAT Tips: Division made easy

**Ravi Handa,an alumnus of IIT Kharagpur suggests three simple rules to find out divisibility of numbers.**

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 Muammar Gaddafi tried using it but as the latest news events show us, he wasn't very effective at gaining from it. 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;-- 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 rules about a few other types of numbers, I suggest that you read on.

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**Image:**Divisibility rules can come in handy while solving problems based on Number Systems

# Funda 1: Check sum of blocks of n

**Example 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.

**Example 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.

** Example 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.

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**Photographs:**Uttam Ghosh

# Funda 2: Check alternating sum of blocks of n

**alternating sum***of blocks of size 'n' needs to be checked (blocks should be considered from the least significant digit, or the right side).

If the alternating sum is divisible by p, then the number is divisible by p.

**Example 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

**Example 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

**Example 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.

** *Alternating sum** is the 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.

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**Photographs:**Uttam Ghosh

# Funda 3: Osculator/seed number method

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**Step 1****:** Figure out an equation such that

p x n = 10m 1

If we have this equation, the osculator/seed number for 'p' will be - or +m.*(-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.

**Example**

Check whether 131537 is divisible by 19 or not.

→ 19 x* *1 = 10 x* *2 - 1 (Seed number is +2)

→ 131537 →13153 + 7 x 2 = 13167→1316 + 7 x 2 = 1330→133 + 0 x 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.

*Author**Ravi Handa** has taught Quantitative Aptitude at IMS for 4 years. An alumnus of IIT Kharagpur where he studied a dual-degree in computer science, he has also written a book on business awareness.*

**Image:**http://im.rediff.com/money/2010/dec/17smart3.jpg

**Photographs:**Uttam Ghosh

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