# Divisibility Rules for 7 , 11 and 13

A guest post by Dr. Cecily Zacharias from Oklahoma city, The United Sates of America. Currently she is an instructor at Oklahoma City Community College

**Rules for divisibility of 7 , 11 and 13**

It is equally good for 11 and 13.

Step 1 Divide into groups of three from the right. 245782 245 782

______________________________________________________________-1 1

Step 2. Write 1,-1,1,-1(alternate 1 and -1) in a row above the number 245 782

( start at the right end and go left)

____________________________________________________________-1 1

Step 3. Divide each Group by 7 (or 11 or 13, whatever the divisor is ) 245 782

0 5

** You can avoid step 2, by simply subtracting first remainder from the second. In this case it will be simply, 5 – 0 = 5*

Step 4. Multiply the corresponding numbers in the top row and bottom row and add 0x -1 + 5x 1 = 5

*** Step 4 can also be avoided. *

Step 5. a. If the sum obtained is zero, The number is divisible by 7 (or 11 or 13 )

b. If the sum is positive, then that is the remainder when we divide the number by 7 ( or 11 or 13 )

c. If the sum is negative, then add 7 (or 11 or 13 ) to get the remainder.

The sum is always less than the divisor.

In the example given, the sum is 5 . Which can be verified.

When 245782 is divided by 7 by long division, the quotient is 35111 and remainder is 5.

If we test for divisor 11, the bottom row will be 3 1

The sum of products of the two rows is -2. since it is negative , add 11 .

So the remainder will be 9

Actual division gives the quotient to be 22343 and remainder 9.

It is the same method for dividing by 13 too.

If you want you can simplify the steps 2 ,to 4 as

Find the remainders in each group and alternately add and subtract the remainders starting from the right. Then use step 5.

**On behalf of all the QuickerMaths.com users, I am highly grateful for her contribution. **

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sorry but the method you have is not very clear can you give clear method of division .

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