Solveeit Logo
Login

Question

Question: Let S<sub>n</sub>=\(\sum_{r = 1}^{n}{r!}\); (n \> 6), then S<sub>n</sub> – 7 \(\left\lbrack \frac{S_...

Let Sn=∑r=1nr!\sum_{r = 1}^{n}{r!}; (n > 6), then Sn – 7 [Sn7]\left\lbrack \frac{S_{n}}{7} \right\rbrack, (where [.] denotes the greatest integer function) is equal to –

A

[n7]\left\lbrack \frac{n}{7} \right\rbrack

B

n! – 7[n7]\left\lbrack \frac{n}{7} \right\rbrack

C

5

D

3

Answer

5

Explanation

Solution

All the number r!, (r ³ 7) will be multiple of 7. So for remainder, consider

S6 = 1 + 2 + 6 + 24 + 120 + 720

Which leaves the remainder 5, when divided by 7

Hence (3) is correct answer.