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Difference between revisions of "2013 IMO Problems/Problem 1"
Line 25: Line 25:
 
==Alternative Solution==
 
==Alternative Solution==
 
We will prove by constructing telescoping product:
 
We will prove by constructing telescoping product:
−
<cmath>\frac{a_2}{a_1}\frac{a_3}{a_2}\frac{a_4}{a_3} \cdot  \frac{a_{k+1}}{a_k} = \frac{a_{k+1}}{a_1} = \frac{\left(a_1+2^{k}-1\right)}{a_1}</cmath>
+
<cmath>\frac{a_2}{a_1}\cdot\frac{a_3}{a_2}\cdot\frac{a_4}{a_3} \cdot  \frac{a_{k+1}}{a_k} = \frac{a_{k+1}}{a_1} = \frac{\left(a_1+2^{k}-1\right)}{a_1}</cmath>
−
where each fraction <math>\frac{a_{i+1}}{a_i}</math> can also be written as <math>\frac{m_i+1}{m_i}</math> for some positive integer <math>m_i</math>
+
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>. Telescoping property implies <math>\sum{\Delta_i}=2^{k}-1</math>. We will show that the set of all <math>\Delta_i</math> can be taken to be a collection of <math>2^j</math> for <math>1\le j \le (k-1)</math>
 +
 
 
--[[User:alexander_skabelin|alexander_skabelin]] 9:24, 11 July 2023 (EST)
 
--[[User:alexander_skabelin|alexander_skabelin]] 9:24, 11 July 2023 (EST)
  

Revision as of 18:25, 11 July 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 by constructing telescoping product: \[\frac{a_2}{a_1}\cdot\frac{a_3}{a_2}\cdot\frac{a_4}{a_3} \cdot \frac{a_{k+1}}{a_k} = \frac{a_{k+1}}{a_1} = \frac{\left(a_1+2^{k}-1\right)}{a_1}\] 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 $m_i$. Telescoping property implies $\sum{\Delta_i}=2^{k}-1$. We will show that the set of all $\Delta_i$ can be taken to be a collection of $2^j$ for $1\le j \le (k-1)$

--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

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