Problem 1351

(Solution by A. Sheremet.) Since $3c! > 2015$, we have $c \ge 6$.

1) For $a = 0$ from the given equation
$$3^a + 2^b + 2015 = 3c! \quad (1)$$
we have $2^b = 3(c! - 672)$, which is impossible.

2) Let $a = 1$. If $c = 6$, then $2^b = 142$ -- there are no solutions. If $c \ge 7$, then (1) implies $2^b + 2018 \ge 7$, or $2^b + 2 \equiv 0 \pmod 7$, which is impossible.

3) Let now $a \ge 2$. Then $3^a \ge 9$, hence from (1) it follows that $2^b \equiv 1 \pmod 9$. Therefore $b \ge 6$.
If $b = 0$, then we have $3^a + 2016 = 3c!$, which is impossible for $c = 6$; for $c \ge 7$ the obtained equality implies $3^a \ge 7$, which is false.
Therefore $b \ge 6$ and from (1) it follows that $3^a \equiv 1 \pmod{16}$ which implies $a \ge 4$. Let $a = 4t$, $b = 6q$. Then (1) becomes
$$81^t + 64^q + 2015 = 3c! \quad (2)$$
The case $c = 6$ leads to $t = q = 1$, i.e., $a = 4$, $b = 6$. The case $c \ge 7$ leads to the congruence $4^t + 1^q + 6 \equiv 0 \pmod 7$, which is impossible. Thus the only solution is $(a, b, c) = (4, 6, 6)$.