You can decide whether a number is divisible by 3 by forming its cross sum, that is the sum of all its decimal digits: It is divisible by three exactly if its cross sum is. Similarly, a three-digit number is divisible by seven if (and only if) the sum of twice its most significant digit plus three times its middle digit plus its least significant digit is divisible by seven.

Can you find a similar scheme for checking divisibility by 33 and 37, for numbers with an arbitrary number of digits?

With the digits being a_0, a_1,... take the following "cross-sum"

a_0 + 10a_1 - 11a_2 +a_3+10a_4-11a_5+...

It can be nicely calculated because no multiplication is required; the "11a_i"-term is simply given by concatenating the digit a_i with itself.

Example:

13140452172 -> 72-11+52-44+40-11+13=111 -> 11-11=0 => divisible by 37

Posted by JLo on 2006-10-25 12:19:36