Let $M=1^n + 2^n + 3^n + 4^n$ We will prove that can only end in two zeros at most by mod $8$ and $n\geq 3$ $M=1+0+(1$ or $3)+0=4$ or $2(mod 8)$ Contradiction! For cases of $n = 1,2$ they do not comply Only one example is missing for $M=...00$ if $n=3$ then $M=100$

parmenides51 wrote: Let $n$ be a positive integer. Find as many as possible zeros as last digits the following expression: $1^n + 2^n + 3^n + 4^n$. First let's notice that if a number ends in at least two zeros, it is 0 in mod 8. (1) If n ≥ 3 then 1ⁿ ≡ 1 (mod 8), 2ⁿ ≡ 0 (mod 8), 3ⁿ ≡ 1 or 3 (mod 8), 4ⁿ ≡ 0 (mod 8), then the sum of the 4 numbers is 2 or 4 in mod 8, and by (1) this is a contradiction. then n ≤ 3 . Now if n = 1 M ends in a zero, if n = 2 ends in a zero, If n = 3 the number ends in 2 zeros. Finally we conclude that the greatest number of zeros is 2.