Question

Find all 4-tuples $(a,b,c,d)$ of natural numbers with $a\ge b\ge c$ and $a!+b!+c!=3^d$ .

Answer 1

Note that if $a>1$ then the left-hand side is even,
$\therefore a=1$.
If b > 2 then 3 divides b! + c! and hence 3 does not divide the left-hand side.
$\therefore b=1$ or $b=2$.
If $b=1$ then $c!+2=3d$, so $c<2$ and
$\therefore d=1$.
If $b=2$ then $c!=3^d-3$.
Note that $d=1$ does not give any solution.
If d > 1 then 9 does not divide c!, so c < 6.
By checking the values for $c=2,3,4,5$ we see that $c=3$ and $c=4$ are the only two solutions.
$\therefore (a,b,c,d)=(1,1,1,1),(1,2,3,2)$ or $(1,2,4,3)$.