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

Correct option is D

Given:

We need to find the smallest natural number n such that (n + 1) × n × (n - 1) × (n - 2) × ... × 3 × 2 × 1 is divisible by 910.

Concept Used:

To determine the smallest n such that the product is divisible by 910, we

need to factorize 910 and ensure that the factorial (n!) contains these

factors.

Formula Used:

Factorize the number 910.

Solution:

Factorize 910:

910 = 2 × 5 × 7 × 13

We need n! to include all these prime factors.

=> The smallest n must be at least 13 to include the factor 13.

Check n = 13:

13! =13× 12 × 11 × 10 × 9 × 8 ×7× 6 ×5× 4 × 3 ×2× 1

13! contains the factors 2, 5, 7, and 13.

∴ 13! is divisible by 910.

Hence, the smallest natural number n is 12.