7
$\begingroup$

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?

$\endgroup$
10
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.