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What is the ones digit of 1^2009 + 2^2009 + 3^2009 + ... + 100^2009?

 May 26, 2021
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Take the entire thing mod $10$.

It's then just $9(1^{2009}+...+9^{2009})\pmod{10}$.

And $1^{2009}=(-9)^{2009}=-9^{2009}\pmod{10}$.

Similarly, $2^{2009}=-8^{2009}\pmod{10}$, etc.

So everything cancels and it becomes $9\cdot 5^{2009}\pmod{10}$. Note that $5^n$ is always $5$ modulo $10$. Then the answer is $9\cdot 5=5\pmod{10}$.

 

(mod 10 is just the remainder when divided by 10)

 May 26, 2021

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