What 3digit number, a product of two distinct 2digit primes has its digits (none of them being prime) in strictly ascending order ?
There are five nonprime digits  1,4,6,8,&9. There are ten combinations of threedigit numbers with digits in ascending order using these digits:
5! / (2!*3!) = 10
These ten 3digit number are:
146, 148, 149, 168, 169, 189, 468, 469, 489, 689
Of these, four can be eliminated because they are even. This leaves:
149, 169, 189, 469, 489, 689
149 is prime, so it has no prime divisors.
169 = 13², so it's prime divisors are not unique.
189 = 3³ * 7, so it does not have only two prime divisors.
489 = 3 * 163, neither of which is two digits.
This leaves 689 = 13 * 53

Posted by hoodat
on 20100417 14:37:10 