Discussion

Question:

Read the following instruction carefully and answer the question that follows:

Expression

$\sum_{n = 1}^{13} \frac{1}{n}$ can also be written as $\frac{x}{13!}$

What would be the remainder if x is divided by 11?

Explanation:

$\sum_{n = 1}^{13} \frac{1}{n} = \frac{x}{13!} ∴ x = \frac{13!}{1} + \frac{13!}{2} + \frac{13!}{3} + . . . + \frac{13!}{11} + \frac{13!}{12} + \frac{13!}{13}$

All the terms in x are divisible by 11 except 13!/11

$\frac{13!}{11} = 1.2.3.4 . . . 10.12.13$

According to Wilson theorem,

$r e m (\frac{(p - 1)!}{p}) = - 1$

$\frac{13!}{11} = 1.2.3.4 . . . 10.12.13 . = 10! (12) (13)$

$∴ r e m (\frac{\frac{13!}{11}}{11}) = \frac{10! (12) (13)}{11} = (- 1) . 1.2 = - 2 = 9$

Hence, option (d).

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