A Sum of Factorials as a Power of 2

Since $2^{d!} \ge 3$ , we have $d \ge 2$ . Start by assuming that $a \ge b \ge c$ . Since $c!$ divides $a! + b! + c!$ , it divides $2^{d!}$ , and hence $c \le 2$ .

If $c=1$ then $a!+b!=2^{d!}-1$ is odd, so $b=1$ . Thus $a! = 2^{d!}-2$ , and so the index of $2$ in $a!$ is $1$ . Thus $a = 2,3$ . It is easy to check that $a=2$ yields $d=2$ , and that $a=3$ is impossible.

If $c=2$ then $a!+b! = 2^{d!}-2$ and $b!$ divides $a!+b!$ , and hence the index of $2$ in $b!$ is at most $1$ , and so $b=2,3$ .

• If $b=2$ then $a! = 4(2^{d!-2}-1)$ , so the index of $2$ in $a!$ is $2$ , which is impossible.
• If $b=3$ then $a! = 8(2^{d!-3}-1)$ , so the index of $2$ in $a!$ is $3$ , so that $a=4,5$ . It is easy to check that neither of these is possible.

Thus the only (ordered) solutions are $(2,1,1,2),(1,2,1,2),(1,1,2,2)$ , making $\boxed{3}$ solutions altogether.

Firstly, since $a!\geq 1$ , we must have $a! + b! + c! \geq 3$ , so $d! \geq 2$ . Thus, the RHS $\equiv 1 \pmod{3}$ .

All unordered triples of residues that have a sum equivalent to 1 mod 3 are $(0,0,1), (0,2,2), (1,1,2)$ . (These will be the residues of $a!$ , $b!$ and $c!$ in mod 3.)

Case 1: $(a!,b!,c!)\equiv (0,0,1) \pmod{3}$

In this case, WLOG $a!\equiv b! \equiv 0 \implies a,b \geq 3$ , $c! \equiv 1 \implies c=1$ . However, $a!$ and $b!$ are even, and $c!=1$ is odd, so the LHS would be odd. This is impossible, so there are no solutions in this case.

Case 2: $(a!,b!,c!)\equiv (0,2,2) \pmod{3}$

In this case, WLOG $a! \equiv 0$ , $b! \equiv c! \equiv 2 \implies b=c=2$ . From this, we get

$a! + 4 = 2^{d!}$

If $a=3$ , we get $2^{d!} = 10$ , which is impossible. Therefore, $a \geq 4$ , so $8 \mid a!$ .

If $d=2$ , we get $2^2 = 4 + a!$ , so $a!=0$ , which is impossible. Therefore, $d \geq 3$ , so $8 \mid 2^{d!}$ .

Notice that the LHS is divisible by 4, but not by 8, whereas the RHS is divisible by 8. Thus, there are no solutions in this case.

Case 3: $(a!,b!,c!)\equiv (1,1,2) \pmod{3}$

From this, we get $a=1$ , $b=1$ and $c=2$ , which we find satisfies, giving $d=2$ . Therefore, our solutions are $(2,1,1,2), (1,2,1,2), (1,1,2,2)$ , so there are 3 solutions.