Find and find again

When $n \geq 10$ then by definition $25 | n!$ and $4|n!$ so $100 | n!$ . Therefore the last two digits of $n!$ is $0$ when $n \geq 10$ . By calculating directly, we get $x = 14$ .

Now to find the last 2 digits of $\sum^{100}_{m=0} m^{14}$ , since $100^{14} \equiv 0 \mod 100$ , we have by using the binomial theorem: $\begin{aligned} \sum^{100}_{m=0} m^{14} &\equiv \sum^9_{r=0} \sum^9_{b=0} (10b+r)^{14} & \mod 100 \\ &\equiv \sum^9_{r=0} \sum^9_{b=0} 140br^{13} + r^{14} & \mod 100 \\ &\equiv \sum^9_{r=0} 45 \times 140r^{13} + 10 \times r^{14} & \mod 100 \\ &\equiv \sum^9_{r=0}10r^{14} & \mod 100 \end{aligned}$

Since last digit of the value we're looking is going to be $0$ , we only need to find the last digit of $\sum^9_{r=0}r^{14}$ which gives us our second to last digit. This doesn't take long at all as $x^5 \equiv x \mod 10 \implies x^{13} \equiv x \mod 10 \implies x^{14} \equiv x^2 \mod 10$ so $\sum^9_{r=0}r^{14} \equiv \sum^9_{r=0}r^{2} \equiv \frac{9 \times 10 \times 19}{6} \equiv 285 \equiv 5 \mod 10$ Hence the last 2 digits of $\sum^{100}_{m=0} m^{14}$ is $50$ .

From $10!$ on, all factorials are divisible by 100, so they won't influence the last two digits of the first sum.

Since \sum_\limits{n = 0}^9 n! = 409114, it immediately follows that $x = 14$ .

Now we notice that $m^{14} = (m^2)^{7}$ .

We also notice, that with

$a \in \{1,2,3,4\}$ and $b \in \{10,20,30,40,50,60,70,80,90\} :$

$(b + a)^2 = b^2 + 2ab + a^2$ and $(b - a)^2 = b^2 - 2ab + a^2$ .

This means that

$(b + a)^{14} \equiv (b^2 + 2ab + a^2)^7 \equiv 7.(b^2 + 2ab + a^2) \mod 100$ and

$(b - a)^{14} \equiv (b^2 - 2ab + a^2)^7 \equiv 7.(b^2 - 2ab + a^2) \mod 100$ .

So $[(b + a)^{14} +(b - a)^{14} ] \equiv 14 a^2 \equiv 2 a^{14} \mod 100.$

Moreover , we can see that

$96^{14} \equiv 7 . (100 - 4)^2 \equiv 7 . (10000 - 400 + 4^2) \equiv 7 . 4^2 \equiv 4^{14} \mod 100$ .

We can deduce similar results for 97,98 and 99, with 3,2,1 (it is left as an exercise for the reader).

We get \begin{aligned} \sum_\limits{m = 0}^{100} m^{14} & \equiv 20.(1^{14} + 2^{14} + 3^{14} + 4^{14}) + \sum_\limits{i = 0}^{9}(10i+5)^{14} & \mod 100 \\ & \equiv 200 + \sum_\limits{i = 0}^{9}(10i+5)^{14} & \mod 100 \\ & \equiv \sum_\limits{i = 0}^{9}(10i+5)^{14} & \mod 100 \\ & \equiv 10 . 5^{14} & \mod 100 \\ & \equiv 70. 25 & \mod 100 \\ & \equiv 50 & \mod 100. \end{aligned}