By Wilson’s theorem, a natural number

*p* > 1 is a prime number iff (

*p* − 1)! + 1 is divisible by

*p*.

∴16! gives a remainder −1 on division by 17.

From the solution above we know that,

*n* gives the same remainder as 31!/17 on division by 17.

= (17 + 14) × (17 + 13) × (17 + 12) × …(17 + 1) × (16 × 15 × 14 × …× 2 × 1)

= (17 + 14) × (17 + 13) × (17 + 12) × …(17 + 1) × 16!

∴ (17 + 14) × (17 + 13) × (17 + 12) × …(17 + 1) × 16! gives the same remainder as

(14! × 16!) on division by 17.

By Wilson's theorem we know that , 16! Gives remainder −1 on division by 17.

∴ 14! × 15 × 16 gives remainder −1 on division by 17.

14! × 15 × 16 gives the same remainder as 14! × −2 × −1 on division by 17.

∴ 14! × 15 × 16 gives the same remainder as 14! × 2 on division by 17.

∴ 14! × 2 gives remainder −1 on division by 17.

∴ 14! will give remainder 8 on division by 17.

∴ 14! × 16! will give remainder 8 × −1 = −8 on division by 17.

∴ 14! × 16! will give remainder 17 − 8 = 9 on division by 17.

∴ 31!/17 will give remainder 9 on division by 17.

∴

*n* will give remainder 9 on division by 17.

Hence, option 4.

*Alternatively*,

The question can be solved without using Wilson's theorem as follows.

We can rewrite the equation as

∴The remainder of

*n* on division by 17 is same as

Let this remainder be

*R*.

All the terms of this expression except the last term are divisible by 17.

∴

*R* is the remainder when (14 × 13 × 12 × … × 2 × 1)

^{2} × 2 × 1 is divided by 17.

(14!)

^{2} × 2 = ((14 × 11)(8 × 2)(13 × 4)(12 × 7)(10 × 5)(6 × 3)(9 × 1))

^{2} × 2

= (154 × 16 × 52 × 84 × 50 × 18)

^{2} × 81 × 2

We can see that all the terms in this expression, except 81 × 2, are of the form 17

*k* ± 1

∴

*R* will be the remainder when 81 × 2 is divided by 17.

∴

*R* = 9

Hence, option 4.