1. The problem states: For an integer $n$ not divisible by 4, find the divisor of the sum $$1^n + 2^n + 3^n + 4^n.$$ 2. We will analyze the sum modulo several divisors to see which divides it regardless of $n$ (not divisible by 4). 3. First, observe the powers of each term modulo 5. Using Fermat's Little Theorem or direct computation for cases: - $1^n mod 5 = 1$ for all $n$. - $2^n mod 5$ cycles with period 4: $2^1=2$, $2^2=4$, $2^3=3$, $2^4=1$, then repeats. - $3^n mod 5$ also has period 4: $3^1=3$, $3^2=4$, $3^3=2$, $3^4=1$. - $4^n mod 5$ has period 2: $4^1=4$, $4^2=1$, then repeats. 4. Since $n$ is not divisible by 4, $n mod 4$ can be 1, 2, or 3. Check sum modulo 5 for these cases: - For $n mod 4 = 1$: sum = $1 + 2 + 3 + 4 = 10 mod 5 = 0$ - For $n mod 4 = 2$: sum = $1 + 4 + 4 + 1 = 10 mod 5 = 0$ - For $n mod 4 = 3$: sum = $1 + 3 + 2 + 4 = 10 mod 5 = 0$ 5. Hence, when $n$ is not divisible by 4, $$1^n + 2^n + 3^n + 4^n$$ is divisible by 5. 6. Checking divisibility by 3 or 7 does not guarantee zero remainder for all such $n$. Final answer: $$\boxed{5}$$