IMO 2013 #1

by Wolstenholme, Oct 27, 2014, 2:14 PM

Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1, m_2, \ldots, m_k$ (not necessarily different) such that
\[ 1+\frac{2^k-1}{n}=\left(1+\frac{1}{m_1}\right)\left(1+\frac{1}{m_2}\right)\dots\left(1+\frac{1}{m_k}\right). \]

Proof:

The result follows immediately from induction and the identities $ \left(1 + \frac{2^{k + 1} - 1}{2n - 1}\right) = \left(1 + \frac{2^{k} - 1}{n}\right)\left(1 + \frac{1}{2n - 1}\right) $ and $ \left(1 + \frac{2^{k + 1} - 1}{2n}\right) = \left(1 + \frac{2^{k} - 1}{n}\right)\left(1 + \frac{1}{2^{k + 1} + 2n - 2}\right) $.

The motivation is pretty natural - just looking at the problem one immediately sees that induction is the way to go and after writing out a few small cases the pattern becomes evident.

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  • This is late, but where is the ARML results post?

    by donot, Aug 31, 2015, 11:07 PM

  • "I am Sam"
    "Sam I am"

    by mathwizard888, Aug 12, 2015, 9:13 PM

  • HW$\textcolor{white}{}$

    by Eugenis, Apr 20, 2015, 10:10 PM

  • Uh-oh ARML practice is Thursday... I should start the homework. :P

    by nosaj, Apr 20, 2015, 12:34 AM

  • Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial :P

    by Wolstenholme, Apr 15, 2015, 10:00 PM

  • How are Chebyshev Polynomials trivial? :P

    by nosaj, Apr 13, 2015, 4:10 AM

  • Are you Sam?

    by Eugenis, Apr 4, 2015, 2:05 AM

  • @Brian: yes, yes I did #whoneedsalgskillz?

    @gauss1181; hey!

    by Wolstenholme, Mar 1, 2015, 11:25 PM

  • hello!!! :D

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  • Hi Wolstenholme did you actually use calc on that tstst problem :o

    by briantix, Aug 2, 2014, 12:25 AM

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