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 18 digits (Posted on 2012-01-12)
Given 5 twos, 4 threes, 6 fives and 3 sevens create a multiplication problem , partial products included. I believe the solution is unique.

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 re(4): a solution + one more Comment 5 of 5 |
(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 3-digit multiplicand exists for only one multiplier in each instance, the multiplier must be 3-digits 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 2012-01-15 04:14:37

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