For how many natural numbers(<=100) is $1111^n +2222^n+3333^n+4444^n$ divisible by 10? For how many natural numbers (0 not included) $n \leq 100$ is $1111^n +2222^n+3333^n+4444^n$ divisible by 10?
I factored out $1111^n$ and got $1111^n(1+2^n+3^n+4^n)$. So $1+2^n+3^n+4^n$ must be divisible by 10. I figured out that this is divisible by 10 for all odd n, but I don't know how to find the other solutions, if any.
 A: If you divide $1^n$, $2^n$, $3^n$, and $4^n$ by 10, each goes through a cycle of remainders:
$1: 1, 1, 1, 1$
$2: 2, 4, 8, 6$
$3: 3, 9, 7, 1$
$4: 4, 6, 4, 6$
So $1^n+2^n +3^n + 4^n$ goes through the cycle of remainders $0, 0, 0, 4$, and thus will be divisible by $10$ whenever $n$ is not divisible by $4$.
A: If $\varphi$ is Euler's totient function, $\varphi(10)=4$. So the exponents are cyclic mod $10$ with a period of $4$. So it suffices to consider $n=1,2,3,4$. Clearly $n=1$ works; for $n=2$, we have $1111^2\cdot 30$; for $n=3$, we have $1111^3\cdot 100$; for $n=4$, we have $1111^4\cdot 354$. So there are $75$ solutions.
A: $$1^n+2^n+3^n+4^n\equiv0\pmod2\text{ for }n\ge0$$
As $2^3\equiv3,2^2\equiv4\pmod5,$
$$1^n+2^n+3^n+4^n$$
$$\equiv1+2^n+(2^3)^n+(2^2)^n$$
$$\equiv\begin{cases}\dfrac{2^{4n}-1}{2^n-1}\equiv 0 &\mbox{if } 2^n-1\not\equiv0\pmod5\iff4\nmid n  \\
4 & \mbox{if } 4\mid n \end{cases}\pmod5$$
$$\implies1^n+2^n+3^n+4^n\equiv0\pmod{[5,2]}\text{ if }4\nmid n$$
A: This can be simplified using the patterns for exponents. $1^n$ always ends in $1$. $2^n$ repeats a pattern where its last digit ends in $2, 4, 8, 6$. $3^n$ repeats a pattern where its last digit ends in $3, 9, 7, 1$. $4^n$ repeats a pattern where its last digit ends in $4, 6$. So adding up our final digits for the four unique cases, we get:
CASE 1: $1+2+3+4 = 10$
CASE 2: $1+4+9+6 = 20$
CASE 3: $1+8+7+4 = 20$
CASE 4: $1+6+1+6 = 14$
Each case corresponding to one of four equivalence classes that the sum can be partitioned into using the equivalence relation that groups the sum by the value each number is being raised to, n, modulus 4.
The cases where the sum of these four least significant digits of the members of the sum are 0 are solutions to our problem. Each of these cases happens 25 times throughout the numbers 1...100 so there are 75 solutions.
