Is there more than one solution to $\frac{1}{K!}+\frac{1}{L!}+\frac{1}{M!}=\frac{1}{N!}$ where $K, L, M, N$ are all natural numbers?
The one solution i came up with was to assume that $K=L=M$, therefore; $\frac{1}{N!} = \frac{1}{K!}+\frac{1}{K!}+\frac{1}{K!}=\frac{3}{K!}$
And from this we can use logical sense to say that $K=3$ and $N=2$, but is there any other solution to this?
Assume without loss of generality that $K \le L \le M$. Then $1/K! \ge 1/L! \ge 1/M!$. Since these three summands add up to $1/N!$ the largest one, $1/K!$, must be at least one-third of $1/N!$. From this we can deduce that either $K = 2, N = 1$ or $K = 3, N = 2$.
In the case $K = 2, N = 1$ you have $1/2 + 1/L! + 1/M! = 1$, so $1/L! + 1/M! = 1/2$. But then either $L = 2$, giving us $1/M! = 0$ which is impossible, or $L = 3$, so $1/L! = 1/6$, and since $1/L! \ge 1/M!$ we can't have $1/L! + 1/M!$ add up to $1/2$.
So we must have $K = 3, N = 2$. Now $1/6 + 1/L! + 1/M! = 1/2$, so $1/L! + 1/M! = 1/3$. Since $1/L!$ is the larger summand it must be at least $1/6$, giving $K = 3, L = 3, M = 3$ as the only solution.