Evaluating $(1^n + 2^n + 3^n + 4^n) \mod 5.$

Question

What will be the reminder of (1ⁿ +2ⁿ +3ⁿ +4ⁿ)/5? Simply (1ⁿ +2ⁿ +3ⁿ +4ⁿ) mod 5=? Where n<=10¹⁰⁰⁰⁰⁰.
My idea: The reminder may be 0 or 1 or 2 or 3 or 4 when a number is divided by 5.
We can easily calculate it when n is too small but when n is too big like n<=10¹⁰⁰⁰⁰⁰ how can we calculate it?

Answer 1

Hint:

If $n$ is odd, $$1^n+4^n$$ is divisible by $1+4$ using Proof of $a^n+b^n$ divisible by $a+b$ when $n$ is odd

If $4\mid n, r^4\equiv1\pmod5$ for $r=1,2,3,4$ generally for $(r,5)=1$

If $n\equiv2\pmod4,$

$$1^n+2^n+3^n+4^n\equiv1^2+2^2+3^2+4^2\pmod5\equiv?\pmod5$$

Answer 2

We can reduce $n$ mod $4$ using Fermat's little theorem,

which indicates that $a^4\equiv1\bmod5$ when $\gcd(a,5)=1$.