Question

Expression $\displaystyle \sum_{n=1}^{13} \frac{1}{n}$ can also be written as $\dfrac{x}{13!}$. What would be the remainder if $x$ is divided by $11$?

• A. 2
• B. 4
• C. 7
• D. 9
• E. None of the above

Hint:

When factorial expressions are taken modulo a prime $p$, most terms vanish unless the denominator cancels $p$. Use Wilson’s theorem $((p-1)!\equiv -1 \pmod{p})$ to simplify such computations.

Answer 1

The Correct Option is D

Step 1: Express $x$.
\[ \sum_{n=1}^{13}\frac{1}{n}=\frac{x}{13!}, ⇒ x=\sum_{n=1}^{13}\frac{13!}{n}. \] Step 2: Work modulo 11.
Since $13!$ is divisible by 11, every term $\tfrac{13!}{n}$ with $n\neq 11$ is a multiple of 11, hence contributes $0 \pmod{11}$.
Only the $n=11$ term survives.
Step 3: Compute the surviving term.
\[ \frac{13!}{11}=1 2 3s 10 12 13. \] Reduce mod 11:
- $12\equiv 1$,
- $13\equiv 2$,
so product $\equiv (1 2s 10) 1 2 \pmod{11}$.
By Wilson’s theorem, $10!\equiv -1 \pmod{11}$. \[ ⇒ \frac{13!}{11}\equiv (-1) 1 2=-2\equiv 9 \pmod{11}. \] \[ \boxed{9} \]