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?

Write the sum as
$$S=(1^5+99^5)+(2^5+98^5)+(3^5+97^5)+.....+ (49^5+51^5)+50^n$$ Then
$100=1+99,=2+98= 3+97,...$ is the common factor of all the baraketed terms. Because $(a^5+b^5)=(a+b)M(a,b)$. Next, $50^5$ is also divisible by 100. So the last two digits in $S$ are $00$.
