A number theory problem by Shivam Agrawal

Note that $\frac{(j-1) \times (j+1)! + j!}{j} \;= \; j \times j! \; = \; (j+1)! - j!$ for any integer $j$ , so the initial sum is $2 + \sum_{j=396}^{1729}\big((j+1)! - j!\big) \; = \; 1730! - 396! + 2!$ so that $a = 287$ , $b = 1333$ and $c=3$ . Let $X = a^{b^c}$ .

Since $a$ is odd, so is $X$ , and hence $X \equiv 1 \pmod{2}$ . On the other hand, $a = 287 \equiv 2 \pmod{5}$ , and hence $a^4 \equiv 1 \pmod{5}$ . Since $b = 1333 \equiv 1 \pmod{4}$ we see that $b^c \equiv 1 \pmod{4}$ , and hence $X \equiv a \equiv 2 \pmod{5}$ . Thus we deduce that $X \equiv \boxed{7} \pmod{10}$ .