Chinese Mathematical Olympiad

The number $\lambda$ is $\frac{1}{1012}$. The minimal such $\lambda$ is $\frac{1}{1012}$. To see that $\lambda <\frac{1}{1012}$ does not work, we take $n=p^{2024}$ for some prime number $p$. Then when writing $n$ as the product of $2023$ integers, at least one of the integers is $p^{\alpha }$ for some $\alpha \ge 2$ (which is not a prime number); then $p^{\alpha }\ge p^2=n^{\frac{1}{1012}}>n^{\lambda }$. Now we prove that $\lambda =\frac{1}{1012}$ satisfies the condition of the problem. To start, we write $n$ as the product of its prime factors $n=p_1{p}_2\cdots p_r$ (allowing some $p_i$'s to be equal), where we order the primes so that $p_1\ge p_2\ge \cdots$. Now starting from $p_1$, let $r_1\ge 1$ be the smallest integer such that $x_1=p_1{p}_2\cdots p_{r_1}>n^{1/2024}$. Note that this $x_1$ satisfies the condition of the problem, namely, if $r_1=1$, $x_1$ is a prime number; otherwise $r_1\ge 2$, then we must have $p_1\le n^{1/2024}$. This means that $p_r\le n^{1/2024}$; so $$x_1=(p_1{p}_2\cdots p_{r-1})p_r\le n^{1/2024}\cdot n^{1/2024}=n^{1/1012},$$ where the first inequality holds by minimality of $r_1$. Next, let $r_2\ge r_1+1$ be the smallest integer such that $x_2=p_{r_1+1}\cdots p_{r_2}>n^{1/2024}$. The same argument as above shows that either $x_2$ is a prime number or $x_2\le n^{1/1012}$. Continue this process. If we run out of $p_i$'s before reaching $x_{2023}$, then set the rest of $x_j$ to be $1$, and then $n=x_1{x}_2\cdots x_{2023}$ gives the needed decomposition. Otherwise, we have constructed $x_1,x_2,\dots ,x_{2022}$ satisfying the condition of the problem (and that each $x_i>n^{1/2024}$ by construction), then we must have $$x_{2023}:=\frac{n}{x_1\cdots x_{2022}}<\frac{n}{(n^{1/2024}{)}^{2022}}=n^{1/1012}.$$ The decomposition $n=x_1{x}_2\cdots x_{2023}$ satisfies the conditions of the problem.

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Alternative solution.

The proof for that $\lambda <\frac{1}{1012}$ does not work is the same as Solution 1. We prove that $\lambda =\frac{1}{1012}$ satisfies the requirement. Let $n=p_1{p}_2\cdots p_r$ be the prime factorization of $n$ (where $p_1\ge p_2\ge \cdots \ge p_r$). For $i=1,2,\dots ,1012$, define $$y_i=\prod _{\begin{array}{c}1\le j\le r\\ j\equiv i(\mathrm{mod}1012)\end{array}}{p}_j.$$ Then $y_1\ge y_2\ge \cdots \ge y_{1012}$ and $y_1{y}_2\cdots y_{1012}=n$. If $y_1\le n^{1/1012}$, then let $x_1,\dots ,x_{2023}$ be $y_1,y_2,\dots ,y_{1012}$ plus $1011$ ones. If $y_1>n^{1/1012}$, since $y_{1012}\le n^{1/1012}$, suppose $y_k>n^{1/1012}\ge y_{k+1}$, then $1\le k\le 1011$. For $1\le i\le k$, we have $$\frac{y_i}{p_i}\le y_{1012}\le n^{1/1012}.$$ Let $x_1,x_2,\dots ,x_{2023}$ be $\frac{y_1}{p_1},\dots ,\frac{y_k}{p_k},y_{k+1},\dots ,y_{1012},p_1,\dots ,p_k$ and $1011-k$ ones.

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Alternative solution.

The proof for that $\lambda <\frac{1}{1012}$ does not work is the same as Solution 1. We prove that $\lambda =\frac{1}{1012}$ satisfies the requirement. Express $n$ as $n=x_1{x}_2\cdots x_{2023}$ such that

$x_1+x_2+\cdots +x_{2023}$ is minimized. (Since there are a finite number of ways to express this, such a representation must exist.) Next, we prove that every $x_i$ greater than $n^{1/1012}$ is a prime number. We use proof by contradiction. Suppose $x_1>n^{1/1012}$ and $x_1$ is not a prime number. Let $x_1=ab$ where $a>1$ and $b>1$, and without loss of generality, assume $a\ge b$. Since $$x_2\cdots x_{2023}=\frac{n}{x_1}<\frac{x_1^{1012}}{x_1}=x_1^{1011}=(\sqrt{x_1}{)}^{2022}\le a^{2022},$$ there exists an $x_i<a$. Assume without loss of generality that $x_2<a$. Express $n$ as $n=a\cdot b\cdot x_2\cdot x_3\cdots x_{2023}$. Since $$(ab+x_2+x_3+\cdots +x_{2023})-(a+b{x}_2+x_3+\cdots +x_{2023})=(b-1)(a-x_2)>0,$$ the expression $n=x_1{x}_2\cdots x_{2023}$ is not the one that minimizes $x_1+x_2+\cdots +x_{2023}$, which is a contradiction! Therefore, $\lambda =\frac{1}{1012}$ satisfies the problem's conditions.

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Alternative solution.

The proof for that $\lambda <\frac{1}{1012}$ does not work is the same as Solution 1. We prove that $\lambda =\frac{1}{1012}$ satisfies the requirement. Let $n=p_1{p}_2\cdots p_k$, where $p_1\ge p_2\ge \cdots \ge p_k$ are all the prime factors of $n$. If $k\le 2023$, then express $n$ as the product of $p_1,p_2,\cdots ,p_k$ and $2023-k$ ones. If $k>2023$, first express $n$ as $n=p_1{p}_2\cdots p_k$. Then, perform the following operation: in each step, select the smallest two factors and combine them into their product, continue this process until only $2023$ factors remain. In each step, as the current number of factors is $\ge 2024$, the product of the smallest two numbers will be $\le n^{2/2024}=n^{1/1012}$. This indicates that every "composite" number formed after each operation is $\le n^{1/1012}$. Combining this with the fact that the initial numbers are primes, it can be concluded that when the operation ends, the $2023$ factors are either prime or do not exceed $n^{1/1012}$. Hence, $\lambda =\frac{1}{1012}$ satisfies the conditions of the problem.