SELECTION and TRAINING SESSION

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

1) For $a=0$ from the given equation $$3^a+2^b+2015=3c!(1)$$ 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 $2^b+2\equiv 0(\mathrm{mod}7)$, which is impossible.

3) Let now $a\ge 2$. Then $3^a\ge 9$, hence from (1) it follows that $2^b\equiv 1(\mathrm{mod}9)$. 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 $3^a\equiv 1(\mathrm{mod}16)$ which implies $a\ge 4$. Let $a=4t$, $b=6q$. Then (1) becomes $$81^t+64^q+2015=3c!(2)$$ The case $c=6$ leads to $t=q=1$, i.e., $a=4$, $b=6$. The case $c\ge 7$ leads to the congruence $4^t+1^q+6\equiv 0(\mathrm{mod}7)$, which is impossible. Thus the only solution is $(a,b,c)=(4,6,6)$.