Find the maximum value of a 9-digit positive integer N such that the product of the digits of N is 9!.

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