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Difference between revisions of "2023 CMO Problems/Problem 1"

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== Problem ==
 
Find the smallest real number <math>\lambda</math> such that any positive integer <math>n</math> can be expressed as the product of 2023 positive integers <math>n=x_1 x_2 \cdots x_{2023}</math>, where for each <math>i \in</math> <math>\{1,2, \ldots, 2023\}</math>, either <math>x_i</math> is a prime number or <math>x_i \leq n^\lambda</math>.
 
Find the smallest real number <math>\lambda</math> such that any positive integer <math>n</math> can be expressed as the product of 2023 positive integers <math>n=x_1 x_2 \cdots x_{2023}</math>, where for each <math>i \in</math> <math>\{1,2, \ldots, 2023\}</math>, either <math>x_i</math> is a prime number or <math>x_i \leq n^\lambda</math>.
  

Revision as of 04:37, 25 May 2024

Problem

Find the smallest real number $\lambda$ such that any positive integer $n$ can be expressed as the product of 2023 positive integers $n=x_1 x_2 \cdots x_{2023}$, where for each $i \in$ $\{1,2, \ldots, 2023\}$, either $x_i$ is a prime number or $x_i \leq n^\lambda$.

Solution 1

1. Let $n=2^{2024}$. Then there exist $x_i \geq 4$ and $\lambda \geq \frac{1}{1012}$.

2. Assume $n=x_1 x_2 \cdots x_{1012} p_1 p_2 \cdots p_k$ where $x_1 \sim x_{1012} \in \mathbb{N}^{+}$and $x_i \leq n^{\frac{1}{1012}}$ with $x_i p_j>n^{\frac{1}{1012}}$. Also, let $p_1 \leq p_2 \leq$ $\cdots \leq p_k$ be primes and $x_1 \leq x_2 \leq \cdots \leq x_{1012}$

We will show that $k \leq 1011$. Suppose otherwise, that $k \geq 1012$. Then \[n \geq x_1^{1012} \cdot p_1^{1012} \Rightarrow x_1 p_1 \leq n^{\frac{1}{1012}}\] which leads to a contradiction. Therefore, the minimum $\lambda$ is: \[\lambda_{\min }=\frac{1}{1012}\] ~moving|szm


See also

2023 CMO(CHINA) (Problems • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All CMO(CHINA) Problems and Solutions
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