What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5? What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5?
My work:
1^5 ends with 1. 
2^5 ends with 2. 
3^5 ends with 3. 
And so on. 
Do I simply add the ending digits to get my answer?

Using Faulhaber's formula:
$$S_{99}=\frac{4a^3-a^2}{4}=a^2\cdot \frac{4a-1}{4}=\frac{9900^2}{4}\cdot \frac{2\cdot 9900-1}{3}=33\cdot 25\cdot 9900\cdot 19799,$$
where:
$$a=\frac{n(n+1)}{2}.$$
Hence, the answer is $00$.
