The presence of a 2 and a 5 would aid greatly in making 2digit subsets composite. Neither can be at the end of the 3digit number, so try 253, 523, 257 and 527.
The ones with 3 allow for 23, a prime.
527 is not itself prime.
That leaves 257.
Any 3digit combination that doesn't include a 2 or doesn't include a 5 will itself be divisible by 3. So the above solution is unique.
Edited on January 22, 2013, 1:46 pm

Posted by Charlie
on 20130122 13:41:36 