Find three different digits that no matter in what order they are arranged, the three digit number formed is the product of two primes.

(In reply to Exhaustive Search via computer program -- spoiler by Charlie)

Well, I used a program to find all the primes less than 1000 then used excel to do the multiplying and a lot of searching the results to find the verious permutations of the digits and found a few more.  Although all the ones I found that Charlie did not had at least one permutation that was a prime itself so the two factors are, of course 1 and itself (which should count since they were not excluded in the problem).  I also came up with quite a few more that had duplicate digits like 1, 1, 9 but since the problem stated, "Find three different digits..." I excluded those. 

Here's what I came up with in addition to Charlie's 1, 7, 8.

1, 3, 4
1, 3, 7
1, 4, 9
1, 7, 9
3, 4, 9
3, 7, 9
4, 7, 9

If anybody wants to see my prime factorizations of the permutations of these, speak now or forever hold you peace.

Edited on March 21, 2005, 8:26 pm

Posted by Sing4TheDay on 2005-03-21 20:22:31