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Difference between revisions of "2013 IMO Problems/Problem 1"

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Therefore, since <math>n</math> was arbitrary, the claim is true for <math>k = m+1</math>, for all <math>n</math>. Our induction is complete and the claim is true for all positive integers <math>n</math>, <math>k</math>.
 
Therefore, since <math>n</math> was arbitrary, the claim is true for <math>k = m+1</math>, for all <math>n</math>. Our induction is complete and the claim is true for all positive integers <math>n</math>, <math>k</math>.
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==Alternative Solution==
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We will prove the claim by constructing telescoping product out of positive integers:
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<cmath>\frac{a_2}{a_1}\cdot\frac{a_3}{a_2}\cdot\frac{a_4}{a_3}\cdot \cdot  \frac{a_{k+1}}{a_k} = \frac{a_{k+1}}{a_1} = \frac{\left(n+2^{k}-1\right)}{n}</cmath>
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where <math>a_1=n</math> and where each fraction <math>\frac{a_{i+1}}{a_i}=\frac{a_{i}+\Delta_i}{a_i}</math> can also be written as <math>\frac{m_i+1}{m_i}</math> for some positive integer <math>m_i</math>. All <math>\Delta_i</math> will be different with <math>\Delta_i=2^j</math>,  <math>0\le j \le (k-1)</math>. <math>\sum_{i=1}^{k}{\Delta_i}=\sum_{i=1}^{k}{2^{i-1}}=2^{k}-1</math>.
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'''Ascend:''' make telescoping product of fractions with a sequence of <math>\Delta_i</math> that increases in magnitude  up to the maximum <math>2^{k-1}</math>. If the maximum <math>\Delta=2^{k-1}</math> is reached go to the descend phase.
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Pull out factors of <math>2</math> (up to the maximum <math>2^{k-1}</math>).
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<math>n=2^j(2z+1)</math>, <math>z+1=2^q(2r+1)</math> etc where <math>j\ge 0</math>, <math>q\ge 0</math> ...
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Construct telescoping as <cmath>\frac{2^j(2z+2)}{2^j(2z+1)}\cdot \frac{2^{j+q+1}(2r+2)}{2^{j+q+1}(2r+1)} ...</cmath>. The corresponding differences <math>\Delta</math> are <math>2^j, 2^{j+q+1}</math> ... Every <math>\Delta_i</math> is bigger then the previous <math>\Delta_{i-1}</math> by at least factor <math>2</math>. Therefore, the biggest needed <math>\Delta=2^{k-1}</math> can be created telescoping at most <math>k</math> fractions. After we constructed the fraction <math>\frac{2^{k-1}(h+1)}{2^{k-1}h}</math> with the biggest needed <math>\Delta=2^{k-1}</math>  we can construct any remaining needed <math>\Delta_i</math> as described below.
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'''Descend:'''
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If we need <math>\Delta=2^{d}</math> where <math>d<k-1</math> we can use as the next telescoping fraction <math>\frac{2^{d}(2^{k-1-d}(h+1)+1)}{2^{d}(2^{k-1-d}(h+1))}</math>. We can construct all the remaining nedded <math>\Delta_i</math> in the decreasing order of their magnitude by repeating the same step.
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--[[User:alexander_skabelin|alexander_skabelin]] 9:24, 11 July 2023 (EST)
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{{alternate solutions}}
  
 
==See Also==
 
==See Also==
 
*[[2013 IMO]]
 
*[[2013 IMO]]
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{{IMO box|year=2013|before=First Problem|num-a=2}}

Latest revision as of 00:30, 19 November 2023

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

$1+\frac{2^k-1}{n}=(1+\frac{1}{m_1})(1+\frac{1}{m_2})...(1+\frac{1}{m_k})$.

Solution

We prove the claim by induction on $k$.

Base case: If $k = 1$ then $1 +\frac{2^1-1}{n} = 1 + \frac{1}{n}$, so the claim is true for all positive integers $n$.

Inductive hypothesis: Suppose that for some $m \in \mathbb{Z}^{+}$ the claim is true for $k = m$, for all $n \in \mathbb{Z}^{+}$.

Inductive step: Let $n$ be arbitrary and fixed. Case on the parity of $n$:

[Case 1: $n$ is even] $1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^{m} - 1}{\frac{n}{2}} \right) \cdot \left( 1 + \frac{1}{n + 2^{m+1} - 2} \right)$

[Case 2: $n$ is odd] $1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^{m}-1}{\frac{n+1}{2}} \right) \cdot \left( 1 + \frac{1}{n} \right)$

In either case, $1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^m - 1}{c} \right) \cdot \left( 1 + \frac{1}{a_{m+1}} \right)$ for some $c, a_{m+1} \in \mathbb{Z}^+$.

By the induction hypothesis we can choose $a_1, ..., a_m$ such that $\left( 1 + \frac{2^m - 1}{c} \right) = \prod_{i=1}^{m} (1 + \frac{1}{a_i})$.

Therefore, since $n$ was arbitrary, the claim is true for $k = m+1$, for all $n$. Our induction is complete and the claim is true for all positive integers $n$, $k$.

Alternative Solution

We will prove the claim by constructing telescoping product out of positive integers: \[\frac{a_2}{a_1}\cdot\frac{a_3}{a_2}\cdot\frac{a_4}{a_3}\cdot \cdot \frac{a_{k+1}}{a_k} = \frac{a_{k+1}}{a_1} = \frac{\left(n+2^{k}-1\right)}{n}\] where $a_1=n$ and where each fraction $\frac{a_{i+1}}{a_i}=\frac{a_{i}+\Delta_i}{a_i}$ can also be written as $\frac{m_i+1}{m_i}$ for some positive integer $m_i$. All $\Delta_i$ will be different with $\Delta_i=2^j$, $0\le j \le (k-1)$. $\sum_{i=1}^{k}{\Delta_i}=\sum_{i=1}^{k}{2^{i-1}}=2^{k}-1$.

Ascend: make telescoping product of fractions with a sequence of $\Delta_i$ that increases in magnitude up to the maximum $2^{k-1}$. If the maximum $\Delta=2^{k-1}$ is reached go to the descend phase. Pull out factors of $2$ (up to the maximum $2^{k-1}$). $n=2^j(2z+1)$, $z+1=2^q(2r+1)$ etc where $j\ge 0$, $q\ge 0$ ... Construct telescoping as \[\frac{2^j(2z+2)}{2^j(2z+1)}\cdot \frac{2^{j+q+1}(2r+2)}{2^{j+q+1}(2r+1)} ...\]. The corresponding differences $\Delta$ are $2^j, 2^{j+q+1}$ ... Every $\Delta_i$ is bigger then the previous $\Delta_{i-1}$ by at least factor $2$. Therefore, the biggest needed $\Delta=2^{k-1}$ can be created telescoping at most $k$ fractions. After we constructed the fraction $\frac{2^{k-1}(h+1)}{2^{k-1}h}$ with the biggest needed $\Delta=2^{k-1}$ we can construct any remaining needed $\Delta_i$ as described below.

Descend: If we need $\Delta=2^{d}$ where $d<k-1$ we can use as the next telescoping fraction $\frac{2^{d}(2^{k-1-d}(h+1)+1)}{2^{d}(2^{k-1-d}(h+1))}$. We can construct all the remaining nedded $\Delta_i$ in the decreasing order of their magnitude by repeating the same step.

--alexander_skabelin 9:24, 11 July 2023 (EST)

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

2013 IMO (Problems) • Resources
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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