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?

$1^5+99^5=(1+99)(1-99+99^2-99^3+99^4)=100(\text{positive number})$ what can you see??
Also 
$2^5+98^5=(100)(2^4-2^3(98)+2^2(98)^2-2(98)^3+98^4)$
My solution is a special case of lab's 
