What is the smallest natural number n such that (n+1) × n × (n−1) × (n−2) × ... × 3 × 2 × 1 is divisible by 910 ?

(n+1)! must be divisible by 910 = 2 × 5 × 7 × 13. The largest prime factor is 13, so (n+1)! must include 13 as a factor, meaning n+1 ≥ 13, i.e., n ≥ 12. For n = 12: 13! is clearly divisible by 910. So smallest n = 12.