There are 362,880 different numbers that use the digits 1 to 9 exactly once (like 187,432,569).

Of these, what percentage are prime numbers?

(In reply to simple guess by Nick Murray)

A simple proof that for any two digits a and b, 10a + b and a + b yeild the same remainder when divded by 9 is 10a + b = 9a + (a + b). This proof can be expanded to any number of digits.

As Nick points out, 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 which yields a remainder of 0 (45, and therefore any number whose digits add up to 45, is evenly divisible by 9).

The proof of Nick claim about divisibility by 3 is a corollary of the proof for divisbility by 9

Posted by TomM on 2002-12-17 03:31:00