0
$\begingroup$

I was looking at the IMO 2013 problems and I was trying to solve the first problem.

Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1$,$m_2$,$...$,$m_k$ (not necessarily different) such that: $$1+\frac{2^k-1}{n}=\left(1+\frac{1}{m_1}\right)\left(1+\frac{1}{m_2}\right)...\left(1+\frac{1}{m_k}\right)$$ However I got stuck and I was very frustrated so I decided to look up for the solution. The solution follows my logic (I was going in the right direction) but it states that:

$$ 1+\frac{2^{m+1}-1}{n}=\left( 1+\frac{2^{m}-1}{\frac{n}{2}}\right)\cdot\left( 1+\frac{1}{n+2^{m+1}-2}\right) $$ Suposing that $n$ is arbitrary fixed and even; and $ m\in\mathbb{Z}^{+} $.

How is the expansion performed?

Here is the link of the answer: https://www.artofproblemsolving.com/Wiki/index.php/2013_IMO_Problems/Problem_1

$\endgroup$
1
  • $\begingroup$ Sorry, my mistake. The question is how do I perform such expansion? $\endgroup$
    – avm
    May 25 '14 at 22:39
0
$\begingroup$

Start by multiplying numerator and denominator by $\frac{1}{2} . $ \begin{align} 1 + \frac{2^m - 1}{n/2} + \frac{1}{n} &= 1 + \frac{2^m - 1}{n/2} + \frac{n + 2^{m+1} - 2}{n(n + 2^{m+1} - 2)} \\ &= 1 + \frac{2^{m+1} - 2}{n} + \frac{1}{n + 2^{m+1} - 2} + \frac{2^{m+1} - 2}{n(n + 2^{m+1} - 2)} \\ &= 1 + \frac{2^{m+1} - 2}{n} + \frac{1}{n + 2^{m+1} - 2} + \frac{2^{m+1} - 2}{n}\frac{1}{(n + 2^{m+1} - 2)}\\ \end{align}

Why we must replace $1$ by the term $ \frac{n + 2^{m+1} - 2}{(n + 2^{m+1} - 2)} $ is still not intuitive, but it gives the result.

$\endgroup$

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.