I'm trying to find the last three digits in number $1^{2013} + 2^{2013} + 3^{2013} + ... + 1000^{2013}$. I started by calculating the remainder for even numbers, since I can present even numbers as $2^i$. By using Euler's theorem and repeated squaring I have calculated that $2^{2013}$ $(mod$ $1000)$ is $192$, $4^{2013}$ $(mod$ $1000)$ is $384$, $8^{2013}$ $(mod$ $1000)$ is $768$ and so on. So I can display remainder of sum of even numbers as $192$ $Σ 2^{(i-1)}$ $(mod$ $1000)$, $i = 0$ to $500$. But I am kind of lost from here on, any ideas on how do I continue?


Note that $$(1000-n)^{2013}+n^{2013}\equiv (-n)^{2013}+n^{2013}\equiv 0\pmod{1000}$$

Therefore $$\sum_{j=1}^{999}j^{2013}=500^{2013}+\sum_{j=1}^{499}[j^{2013}+(1000-j)^{2013}]\equiv 0\pmod{1000}$$

Then, the three last digits of the sum are zeros.

  • $\begingroup$ we get also the last four digits of the sum are zeros $\endgroup$ – Dr. Sonnhard Graubner Nov 23 '14 at 19:24

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