+0

# Units Digit

0
65
1

What is the ones digit of 1^2009 + 2^2009 + 3^2009 + ... + 2009^2009 + 2010^2009 + 2011^2009 + 2012^2009?

Jul 28, 2021

Taking mod 10, it's just $201(1^{2009}+2^{2009}+...+9^{2009})+1^{2009}+2^{2009}\equiv 1^{2009}+...+9^{2009}+1+2^{2009}\pmod{10}$. Now note that $9^{2009}=-1^{2009}$, etc. so it is just $5^{2009}+2^{2009}+1\pmod{10}$. But $5$ is always $5$ mod 10, and $2$ cycles as $2,4,8,6$. Hence the answer is $8$.