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Calculation

This code calculates $\sum_{n=1}^M n!\ \text{mod}\ 13$ where $M$ is the highest integer for which $M! < 13^{1000}$ .

Because $n!\ \text{mod}\ 13 = 0$ for $n \geq 13$ and $13! < 13^{1000}$ , this is equal to $\sum_{n=1}^{12} n!\ \text{mod}\ 13 = 1 + 2 + 6 + 11 + 3 + 5 + 9 + 7 + 11 + 6 + 1 + 12 = \boxed{74}.$

Brief explanation

Let $N$ be the number of times the main loop has been run. Then $A[]$ is the representation of $N$ in base 13, and $B[]$ is the factorial representation of $N$ . That is, $N = \sum_{k = 0}^{3999} A[k]\ 13^k\ \ \ \ \text{with}\ \ \ 0 \leq A[k] < 13; \\ N = \sum_{k = 0}^{3999} B[k]\ (k+1)!\ \ \ \ \text{with}\ \ \ 0 \leq B[k] \leq k+1.$

The condition $\ell > -1$ will be true precisely if in representation $B[]$ a digit becomes non-zero that has not been non-zero before. This happens precisely with $N = 1!, 2!, 3!, \dots$ . (The variable $k$ keeps track of the last digit that turned non-zero.)

Whenever $\ell > -1$ , the least significant digit in the base-13 representation $A[]$ is added to the running total $m$ .

The conclusion is, therefore, that we add $N$ mod 13 for $N = 1!, 2!, 3!, \dots$ The conclusion above follows easily.