Another Remainder Question

$x=13!(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{13})$

$x=(\frac{13!}{1}+\frac{13!}{2}+\frac{13!}{3}+...+\frac{13!}{11}+\frac{13!}{12}+\frac{13!}{13})$

In the above expression, there are $13$ terms and except $\frac{13!}{11}$ all other terms are divisible by $11$ ,leaving no remainder. So the remainder would depend solely on $\frac{13!}{11}=10! \times 12 \times 13$

By Wilson's theorem $10! \equiv -1 \pmod{11}$

Also $12 \equiv 1 \pmod{11}$

$13 \equiv 2 \pmod{11}$

So $x \equiv -1 \times 1 \times 2 \equiv -2 \pmod{11}\equiv 9 \pmod{11}$

Hance the remainder is $\boxed{9}$ .

A solution without appealing to Wilson: $x = \frac{13!}{1} + \frac{13!}{2} + \cdots + \frac{13!}{13},$ and the only term that does not contain the factor 11 is $13!/11$ , where it is divided out at the bottom. So we look at the remainder of $1\cdot 2\cdots 9\cdot 10\cdot 12\cdot 13,\ \ \ \ \text{when divided by 11}.$ Now we may replace each factor by another number with the same remainder after division by 11; here we have $2\cdot 3\cdot 4\cdot (-6)\cdot 6\cdot (-4)\cdot (-3)\cdot (-2)\cdot (-1)\cdot 1\cdot 2 \\ = (2\cdot 6)\cdot (-2\cdot -6)\cdot (3\cdot 4)\cdot (-3\cdot -4)\cdot (-1)\cdot 2 \\ = 12^4\cdot (-2),$ which we further reduce to $(-1)^4\cdot (-2) = -2/$ which has remainder $\boxed{9}$ after division by 11.