What is the reminder when $1^1+2^2+3^3+\ldots+50^{50}$ is divided by 8 I have a sum that goes as follows $$P=1^1+2^2+3^3+\ldots +49^{49}+50^{50}.$$
The question is to find the reminder when $P$ is divided by 8.
Inorder to find this, I seperated $P$ into $P_1$ and $P_2$ where $$P_1=2^2+4^4+\ldots+50^{50},$$
$$P_2=1^1+3^3+\ldots+49^{49}.$$
And I realised that $P_1 \text{ mod } 8 = 4 $ and $ P_2 \text{ mod } 8 = 1$.
But I want to know if there is an easier, straightforward method for finding the answer.
 A: This is still kind of brute force, but it's much more optimized, and it's a solution you can easily generalize to $\sum_{k=1}^{n} k^k \pmod{m}$ for higher values of $n$.
By the rules of modular arithmetic, when exponentiating, you can reduce the base modulo 8, and then break the terms into groups of 8. So, instead of writing $1^1 + 2^2 + \cdots + 50^{50}$, you can write $(1^1 + 2^2 + \cdots + 8^8) + (1^9 + 2^{10} + \cdots + 8^{16}) + \cdots +  (1^{41} + \cdots + 8^{48}) +  1^{49} + 2^{50}$.
Further rearranging, we can write this as
$$ (1^1 + 1^9 + \cdots + 1^{49}) + \cdots + (8^8 + 8^{16} + \cdots + 8^{48}). $$
Now we can analyze this by casework, into 8 cases.

*

*$1^{n}$ = 1 for all n, so the first subcase sums to 7.

*$2^2 = 4$, and for any $n\geq 3$, $2^n$ is divisible by 8. Hence this second subcase is $4 \pmod{8}$.

*Note that $3^8 = 1 \pmod{8}$. Hence every term is the same mod 8. This contributes $3 \cdot 6 = 18 = 2 \pmod{8}$.

*Writing $4^n$ as $2^{2n}$, we see that all terms are divisible by 8.

*This is the same as the third case; $5^8 = 1 \pmod{8}$. Hence every term is the same mod 8, contributing $5 \cdot 6 = 30 = 6 \pmod{8}$.

*For all $n\geq 3$, $6^n$ is divisible by 8. Hence this case contributes nothing modulo 8.

*$7^8 = 1 \pmod{8}$. Applying the same logic, we have $7\cdot 6=42 = 2 \pmod{8}$.

*This last term contributes nothing because $8 | 8^n$ for all $n\geq 1$.

Adding up all the values, we get $7+4+2+6+2=21 = 5 \pmod{8}$. This is our answer.
