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Nine Factorial Nuance (Posted on 2014-12-09) Difficulty: 3 of 5
Find the maximum value of a 9-digit positive integer N such that the product of the digits of N is 9!.

No Solution Yet Submitted by K Sengupta    
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Solution answer Comment 2 of 2 |
9! = 9×8×7×6×5×4×3×2×1.

Retaining the largest digits, 9 and 8, that are composed of the
factors (3×3) and (2×2×2), respectively, the factors of the nine
multiplicands are 9×8×7×(3×2)×5×(2×2)×3×2×1(×1×1×...×1).

Regrouping into nine groups with the smaller factors combined to form the higher digits:
9×(3×3)×8×(2×2×2)×7×5×2×1(×1×1×...×1).
= 9×9×8×8×7×5×2×1×1.

Thus the 9-digit positive integer is 998875211.

  Posted by Dej Mar on 2014-12-09 14:14:59
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