*Wednesday, January 22nd, 2014*

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 2^{n} and 5^{n}. 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 2^{n} or 5^{n}, then the number is divisible by 2^{n} or 5^{n} and vice versa. For details about other numbers, I suggest that you read on.

**Funda 1:**

** **For checking divisibility by ‘p’, which is of the format of 10^{n} – 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 10
^{1}– 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 10
^{2}– 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 10
^{3}– 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:**

** **For checking divisibility by ‘p’, which is of the format of 10^{n} + 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 10
^{1}+ 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 10
^{2}+ 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 10
^{3}+ 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: ****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.