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 threedigit 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, a_2,... calculate the
vswitchscross sum a_0+10*a_1+a_2+10*a_3+... and check if it is divisible by 33. Repeat if necessary until you arrive at a two digit number. Example:
8419752 > 8+41+97+52 = 198 > 1+98=99 > divisible by 33
8419757 > 8+41+97+57 = 203 > 3+2=5 > leaves remainder 5 when divided by 33, i.e. not divisible.

Posted by JLo
on 20061025 11:40:41 