Monday, December 12, 2011

divisibility by 7

Hi, recently i came across a problem where it was required to check if a number is divisible by 7 . Though quiet an elementary task, i think its not very commonly known. This is what i found.
Test 1:
Reverse the number . Mulitply each digit of this reversed number by the digits 1(=1%7),3(=10%7),2(=100%7),6(=1000%7),4(=10000%7),5(=100000%7) consequtively till all digits are covered. Sum the products. If this value is divisible by 7 then the original number is too.
Actually these numbers are the cyclic remainders of division by 7.
Test 2:
Take the last digit of the number and subtract it from the number formed by the remaining digits.If this value is divisible by 7 then the original number is too. We can do it recursively to reduce the number .

Hope that helps.

No comments:

Post a Comment