Divisibility Rules for 7 , 11 and 13

Posted: 19 Oct 2010 11:58 PM PDT

A guest post by Cecily Zacharias from Oklahoma city, The United Sates of America

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.