Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Corresponding terms from the sequence $1,2,3,4,5...$ and $2^1,2^2,2^3,2^4,2^5,...$, are multiplied, creating the sequence $1\times 2^1,2\times 2^2,3\times 2^3,4\times 2^4,5\times 2^5...$. Let $A$ be the sum of the first $2011$ numbers in the new sequence. Find the remainder when $A$ is divided by $1000$

My solution:

$2S-S=2010 \cdot 2^{2012}+2$. Then $2^{2012} \equiv 2^{12} \mod{125} \equiv 96 \mod{125}$ $2010 \cdot 2^{2012} \equiv 10 \cdot 96 \mod{1000} \equiv 960 \mod{1000}$ Add 2 and the answer is $962$

share|cite|improve this question
Yes that's right. – anon Jul 21 '12 at 3:13

I'm answering this so the question can be marked as answered. If you ask such "am I right?" questions in the future you might want to include some more of your thinking for the benefit of the reader.

The solution is correct. The step $2^{2012}\equiv2^{12}\bmod125$ uses Euler's theorem together with $\phi(5^3)=5^3-5^2=100$. The remainder $2^{2012}\equiv96\bmod1000$ is being computed by computing the remainders modulo the prime powers $5^3=125$ and $2^3=8$ and noting that the remainder $\bmod8$ is $0$. Alternatively, you could have used $\phi(5^3\cdot2^3)=(5^3-5^2)(2^3-2^2)=400$ directly to get $2^{2012}\equiv2^{12}\equiv96\bmod1000$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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