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Find the number of 4-tuples (a,b,c,d) of natural numbers with $a \le b \le c$ and $a! + b! + c! = 3^d$

Discussion:

The basic idea is: factorial function is faster than exponential function in the long run. Note that all three of a, b, c cannot be larger than 3; then the left side will be divisible by 4 but right side is not. Hence the possible values of a are 1, 2 and 3.

If a=1, then b and c both cannot be greater than or equal to 3 (then the left hand side is not divisible by 3). So we have the following cases:

b=1 or b=2

If b =1 then we have $1! + 1! + c! = 3^d$ Surely c! is 1 mod 3 (otherwise the left hand side is not divisible by 3). Then the possible values of c is 1. And indeed $1! + 1! + 1! = 3^1$ fits into our equation. Hence (1, 1, 1, 1) is a solution and for b=1 there are no other.

If b=2 then we have $1! + 2! + c! = 3^d$. Surely c! is 0 mod 3. Hence possible values of c are 3, 4, 5, …

c=3 and c=4 furnish specific solutions as $1! + 2! + 3! = 3^2$ and $1! + 2! + 4! = 3^3$ . Hence (1, 2, 3, 2) and (1, 2, 4, 3) are two solutions.

Can c be greater than 4? Surely c cannot be 5 since $1! + 2! + 5! = 123$ is not a power of 3. From c=6 onward we argue $1! +2! + c! = 3^d$ implies $3(1 + \frac {c!}{3} ) = 3^d$ implies $1 + \frac {c!}{3} = 3^{d-1}$ . Since c is greater than 5, c! will contain atleast two 3’s in it’s prime factorization (and d will be greater than 4 as 5! = 120 > 81 ) . Hence $\frac {c!}{3}$ is divisible by 3.
Thus in the equation $1 + \frac {c!}{3} = 3^{d-1}$ left hand side is 1 mod 3 (that is produces 1 as remainder when divided by 3) and right hand side is 0 mod 3. Hence no solution.

Hence there are exactly 3 solutions.

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