A question about sums and factorials

Question

Consider the sum $S=x!+\sum_{i=0}^{2013}i!$, where $x$ is a one-digit nonnegative integer. How many possible values of $x$ are there so that S is divisible by 4?

Answer 1

HINT:

$$\sum_{i=0}^{2013}i!\equiv\sum_{i=0}^3i!\pmod 4\equiv 0!+1!+2!+3!=10 $$

Answer 2

If we expand the sum, we get that $$0!+1!+2!+3!+4\left(\frac{4!}{4}+\frac{5!}{4}+...+\frac{2013!}{4}\right)=10+4\left(3!+\frac{5!}{4}+...+\frac{2013!}{4}\right).$$

We now have determined that $x!+10$ must be congruent to $0$ mod $4$.

Answer 3

HINT: Note that $k!$ is divisible by $4$ for all $k\ge 4$, so

$$\sum_{k=0}^{2013}k!=0!+1!+2!+3!+\sum_{k=4}^{2013}k!=10+4n=2+4(n+2)$$

for some integer $n$. Thus, $S$ is divisible by $4$ if and only if $x!+2$ is divisile by $4$.

Answer 4

Hint:

All the factorials from $4!$ to $2013!$ are divisible by $4$, so the sum of them is also divisible by $4$.

$0!+1!+2!+3!=10$.

You need to solve that $x! +10$ is divisible by $4$.

Good luck!