What is the smallest natural number n such that (n+1) \times n \times (n-1) \times (n-2) \times \dots 3 \times 2 \times 1 is divisible by 910?

Explanation

The product is (n+1)!. Prime factorization of 910 = 2 \times 5 \times 7 \times 13. For the factorial to be divisible by 910, it must contain the largest prime factor, 13. Thus, n+1 \geq 13 \implies n \geq 12.

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