# Solution to Puzzle #70 - Divisibility Test

I don’t normally post solutions, but I think this is quite mathematically interesting.

The problem was to find an easy test for divisibility by 7, 11 or 13, by reducing a number of however many digits to a number with only 3 digits that would have the same divisibility by 7, 11 and 13 as does your original number.

Utilising the fact that 7x11x13 = 1001, you can easily ‘cast out’ 1001s via the following method:

Chop the number into chunks of 3 digits, starting from the right hand side (you may of course end up with one or two digits at the left hand side). If the number is already shown with ‘thousand separator’ commas, this is done for you:

8363859572634 becomes 8,363,859,572,634

Alternately add and subtract your three digit chunks. Eg

8-363+859-572+634 = 566

If the resulting number is negative, just ignore its minus sign.

If the resulting number is still more than 3 digits long, just repeat the procedure.

Not only will the 3 digit number be divisible by 7, 11 or 13 if and only if the original number was, but if it isn’t divisible its remainders will be the same as the original number!