The presence of a 2 and a 5 would aid greatly in making 2-digit subsets composite. Neither can be at the end of the 3-digit 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 3-digit 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
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Posted by Charlie
on 2013-01-22 13:41:36 |