<s>[INST] Solving $a!+b!+c!=3^d$ The question is to find all tuples $(a,b,c,d)$ of natural numbers $c\geq b$ and $ b \geq a $ and $a!+b!+c!=3^d$. I am finding difficulty in establishing relation between $a$, $b$, $c$, and $d$. I see that a!,b!,c! are even.How can their sum be odd? Please help.
 [/INST] When $a \le b \le c$ we get

$$
a! + b! + c! = a! \Big( 1 + b!/a! \big( 1 + c!/b! \big) \Big) = 3^d. \tag 1
$$

If $a \ge 2$ then there are no solutions.
If $b \ge 3$ then there are no solutions.
So we can only get

$$
1 + 2 + c! = 3^d, \tag 2
$$

thus

$$
c! = 3 \big( 3^{d-1} - 1 \big) \tag 3
$$

But $3$ divides $c!$, so $3 \le c$.
But $3$ does not divide $3^{d-1} - 1$, so $c < 6$.
So

$$
2 < c < 6. \tag 4
$$

Verification

$$
3! = 3 \Big( 3^{d-1} - 1 \Big) \Rightarrow 3^d = 9,\\ \tag 5
4! = 3 \Big( 3^{d-1} - 1 \Big) \Rightarrow 3^d = 27,\\
5! = 3 \Big( 3^{d-1} - 1 \Big) \Rightarrow 3^d = 123.\\
$$

So the only solutions are

$$
\bbox[16px,border:2px solid #800000] {
\begin{array}{c}
(1,2,3) \Rightarrow 1! + 2! + 3! = 3^2\\\\
(1,2,4) \Rightarrow 1! + 2! + 4! = 3^3
\end{array} } \tag 6
$$

</s>