Evaluate the remainder when $\sum_{k=1}^{30}(-2)^k\cdot\frac{30!}{k}$ is divided by $992=32\cdot31$?
I believe I have a solution to this problem, but it requires some computations with large numbers, and am looking to see if there are any shortcuts that cut the necessary computations (or any nicer/faster solutions altogether).
Here is what I did:
Using CRT, we just need to work mod $32$ and mod $31$. Working mod $32$, we can see easily see that the sum is $0$, because $v_2(30!)=26>4+5=9$, where $9$ is an upper bound in the number of powers of $4$ the denominator can have, and $5=v_2(32)$. (This bound is very weak as $4$ clearly can't be attained, but it doesn't matter).
Now we work mod $31$. This is pretty tricky. We first apply the (somewhat known) relation $\frac{1}{k}\equiv(-1)^{k-1}\frac{1}{p}\binom{p}{k}\pmod{p}$. (More about this can be found here). Then the sum simplifies to $-\frac{1}{31}\sum_{k=1}^{30}2^k\binom{31}{k}=-\frac{1}{31}[(2+1)^{31}-2^{31}-1] =-\frac{3^{31}-2^{31}-1}{31} \pmod{31}$.
This can be evaluated by letting the numerator be $31a+b$ with $0\leq a,b<31$ (which is essentially working mod $31^2$), and then evaluating $a\pmod{31}$, even though this requires a bit of tedious calculations (as $b$ clearly must be $0$).
Any ideas? All thoughts/comments will be appreciated.