> ... Of course it would be a number involving 7 and 13. We didn't have a > rule for 7 and 13 was outside of his multiplication facts. >
It's not hard to develop a rule for seven. Maybe he can do it if you first ask him why the rule for 3 works, and then show him if he can't figure that out himself. [Hint: 725 = 7(100) + 2(10) + 5 = 7(99 + 1) + 2(9 + 1) + 5.]
For six-digit numbers, n, the rule is this: n is divisible by 7 if, and only if, the sum of the ones-digit, three times the tens-digit, two times the hundreds-digit, six times the thousands-digit, four times the ten-thousands-digit, and five times the hundred-thousands-digit is divisible by seven. That is
For numbers of more than six digits, use the fact that the coefficients are periodic, with period of length six.
Divisors greater than ten are a little more difficult, because they're most efficiently done using negative numbers. For example, a number is divisible by eleven iff the sum of its digits *with alternating sign* is divisible by eleven.