Solve on$ \,\,\Bbb N$ :
$$ 2^m+3^n+5^p+7^q+11^r=s!$$
In my attempt to solve this equation. I found $(0,1,0,0,0,3)$ and $(2,0,0,,1,1,4)$ and some more solutions. I tried to reduce it using congruent, I failed.Question proposed by Jalil Hajimir. $$\Bbb N=\{0,1,2,3,...\}$$
Edit: I tried $\mod 7$ and also $\mod 2\times3\times5\times7\times11$, I know it does not have any solution beyond $s=11$ " as a friend mentioned".
$$\forall s>2,~2^m+5^p+7^q+11^r\equiv 0\pmod 3\implies 2^m+2^p+2^r+1\equiv 0\pmod 3$$ implying 1 or all of $m,p,r$ are odd.
If $$m\neq 1$$:
$$\forall s>3,~3^n+5^p+7^q+11^r\equiv 0\pmod 4\implies 3^n+3^q+3^r+1\equiv 0\bmod 4$$ implying 1 or all of $n,q,r$ are odd.
$$\forall s>4,~2^m+3^m+7^q+11^r\equiv 0\pmod 5\implies 2^m+(-2)^n+2^q+1\equiv 0\bmod 5 $$ in the case that $n$ is even, we get that either 2 give back additive inverses, leading to the other being 2 mod 4, or we have two of $m,n,q$ being equal mod 4, forcing the other exponent into a certain value based on the value,1,2,3,4 mod 5 the others land on.
Most of this is simply using $s!\equiv 0\pmod s$ casting out constants that appear, and attempting to equate the rest. it's a fairly general method.