Let $A$, $B$, $C$ be integers such that $0 \le A, B, C \le 9$. Find all the solutions for $$A\times B\times C = A! + B!+ C!$$
I have tried some values for $A$, $B$, and $C$ and found a solution: $A=4$, $B=3$, $C=3$.
Is there a way to solve this without using a computer to try every combination possible?
We may suppose that $A\le B\le C$. Then $ABC\le C^3$. But if $C\ge 6$, then $C!>C^3$, so $$ABC<A!+B!+C!$$ Hence $C\le 5$. And if $C=5$, then we have $$5AB=A!+B!+120$$ If $A\le 4$, then the LHS is $\le 100$; and if $A=5$, then the RHS is $>125$. Both cases lead to a contradiction; therefore $C$ can't be $5$.
Hence $C\le 4$. And you seem to have covered all those cases.
I don't think you can do it without some testing, but you don't have to try every possibility. We can assume $$A\geq B\geq C$$
The right-hand side is greater than $A!$. Dividing through by $A$ gives $BC>(A-1)!$, so certainly $81>(A-1)!$ and $A<6$.
What if $A=5$? Then we have, as above $$BC>24$$ and $C\leq B\leq A=5$, so $A=B=C=5$, but this isn't a solution.
So now we know $A\leq4$, and we proceed in the same way.
Partial answer to limits the possibilities.
Since $A, B, C\leq 9$, then $ABC\leq 9^3=729$. Since $6!=720$, then $A, B, C \leq 6$.
We could continue with the same argument. $$ABC\leq6^3=216 \text{ and }5!=120\implies A, B, C \leq 5.$$ This doesn't solve it but it limits the possibilities.
