Given 5 twos, 4 threes, 6 fives and 3 sevens create a multiplication problem , partial products included.
I believe the solution is unique.
(In reply to
re(3): a solution + one more by Ady TZIDON)
The only digit that results in one of the prime digits {2, 3, 5, 7} when multiplied by two prime digits is 5. Iterating through the possible four digit numbers in producing partial products that end in 5 and are all prime digits gives only four results: {2275, 2325, 2775, 3775} which are the respective products of
{325 x 7, 775 x 3, 555 x 5, 755 x 5}. As the 3digit multiplicand exists for only one multiplier in each instance, the multiplier must be 3digits of the same digit. The only one that results in a product of only prime digits is 325 x 777.
325
777
2275
2275
2275
252525
2's : 10
3's : 1
5's : 7
7's : 6

Posted by Dej Mar
on 20120115 04:14:37 