What 3-digit number, a product of two distinct 2-digit primes has its digits (none of them being prime) in strictly ascending order ?

There are five non-prime digits - 1,4,6,8,&9.  There are ten combinations of three-digit numbers with digits in ascending order using these digits:

5! / (2!*3!) = 10

These ten 3-digit 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 2010-04-17 14:37:10