2010 IMO Shortlist N1

Find the least positive integer $n$ for which there exists a set $\left\{s_{1}, s_{2}, \ldots, s_{n}\right\}$ consisting of $n$ distinct positive integers such that $$\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right)=\frac{51}{2010}$$ $\mathbf{N} 1^{\prime}$. Same as Problem N1, but the constant $\frac{51}{2010}$ is replaced by $\frac{42}{2010}$. (Canada) Answer for Problem N1. $n=39$. Solution for Problem N1. Suppose that for some $n$ there exist the desired numbers; we may assume that $s_{1}<s_{2}<\cdots<s_{n}$. Surely $s_{1}>1$ since otherwise $1-\frac{1}{s_{1}}=0$. So we have $2 \leq s_{1} \leq s_{2}-1 \leq \cdots \leq s_{n}-(n-1)$, hence $s_{i} \geq i+1$ for each $i=1, \ldots, n$. Therefore $$\begin{aligned} \frac{51}{2010} & =\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right) \\ & \geq\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \ldots\left(1-\frac{1}{n+1}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{n}{n+1}=\frac{1}{n+1} \end{aligned}$$ which implies $$n+1 \geq \frac{2010}{51}=\frac{670}{17}>39$$ so $n \geq 39$. Now we are left to show that $n=39$ fits. Consider the set $\{2,3, \ldots, 33,35,36, \ldots, 40,67\}$ which contains exactly 39 numbers. We have $$\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{32}{33} \cdot \frac{34}{35} \cdots \frac{39}{40} \cdot \frac{66}{67}=\frac{1}{33} \cdot \frac{34}{40} \cdot \frac{66}{67}=\frac{17}{670}=\frac{51}{2010}$$ hence for $n=39$ there exists a desired example. Comment. One can show that the example (1) is unique. Answer for Problem N1' ${ }^{\prime} n=48$. Solution for Problem N1'. Suppose that for some $n$ there exist the desired numbers. In the same way we obtain that $s_{i} \geq i+1$. Moreover, since the denominator of the fraction $\frac{42}{2010}=\frac{7}{335}$ is divisible by 67 , some of $s_{i}$ 's should be divisible by 67 , so $s_{n} \geq s_{i} \geq 67$. This means that $$\frac{42}{2010} \geq \frac{1}{2} \cdot \frac{2}{3} \cdots \frac{n-1}{n} \cdot\left(1-\frac{1}{67}\right)=\frac{66}{67 n}$$ which implies $$n \geq \frac{2010 \cdot 66}{42 \cdot 67}=\frac{330}{7}>47$$ so $n \geq 48$. Now we are left to show that $n=48$ fits. Consider the set $\{2,3, \ldots, 33,36,37, \ldots, 50,67\}$ which contains exactly 48 numbers. We have $$\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{32}{33} \cdot \frac{35}{36} \cdots \frac{49}{50} \cdot \frac{66}{67}=\frac{1}{33} \cdot \frac{35}{50} \cdot \frac{66}{67}=\frac{7}{335}=\frac{42}{2010}$$ hence for $n=48$ there exists a desired example. Comment 1. In this version of the problem, the estimate needs one more step, hence it is a bit harder. On the other hand, the example in this version is not unique. Another example is $$\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{46}{47} \cdot \frac{66}{67} \cdot \frac{329}{330}=\frac{1}{67} \cdot \frac{66}{330} \cdot \frac{329}{47}=\frac{7}{67 \cdot 5}=\frac{42}{2010} .$$ Comment 2. N1' was the Proposer's formulation of the problem. We propose N1 according to the number of current IMO.

找到最小正整数 $n$，其中存在由 $n$ 个不同正整数组成的集合 $\left\{s_{1}, s_{2}, \ldots, s_{n}\right\}$，使得 $$\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right)=\frac{51}{2010}$$ $\mathbf{N} 1^{\prime}$。与问题 N1 相同，但常数 $\frac{51}{2010}$ 被 $\frac{42}{2010}$ 替换。 (加拿大)问题N1的答案。 $n=39$。问题N1的解决方案。假设对于某些 $n$ 存在所需的数字；我们可以假设 $s_{1}<s_{2}<\cdots<s_{n}$。当然$s_{1}>1$，否则$1-\frac{1}{s_{1}}=0$。所以我们有 $2 \leq s_{1} \leq s_{2}-1 \leq \cdots \leq s_{n}-(n-1)$，因此对于每个 $i=1, \ldots, n$ 都有 $s_{i} \geq i+1$。因此 $$\begin{aligned} \frac{51}{2010} & =\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right) \\ & \geq\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \ldots\left(1-\frac{1}{n+1}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{n}{n+1}=\frac{1}{n+1} \end{aligned}$$ 这意味着 $$n+1 \geq \frac{2010}{51}=\frac{670}{17}>39$$ 因此 $n \geq 39$。现在我们需要证明 $n=39$ 适合。考虑集合 $\{2,3, \ldots, 33,35,36, \ldots, 40,67\}$，它恰好包含 39 个数字。我们有 $$\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{32}{33} \cdot \frac{34}{35} \cdots \frac{39}{40} \cdot \frac{66}{67}=\frac{1}{33} \cdot \frac{34}{40} \cdot \frac{66}{67}=\frac{17}{670}=\frac{51}{2010}$$ 因此对于 $n=39$ 存在一个所需的示例。评论。可以证明例子(1)是唯一的。问题 N1' 的答案 ${ }^{\prime} n=48$。问题N1'的解决方案。假设对于某些 $n$ 存在所需的数字。以同样的方式我们得到$s_{i} \geq i+1$。此外，由于分数 $\frac{42}{2010}=\frac{7}{335}$ 的分母可以被 67 整除，所以 $s_{i}$ 中的一些应该可以被 67 整除，所以 $s_{n} \geq s_{i} \geq 67$。这意味着 $$\frac{42}{2010} \geq \frac{1}{2} \cdot \frac{2}{3} \cdots \frac{n-1}{n} \cdot\left(1-\frac{1}{67}\right)=\frac{66}{67 n}$$ 这意味着 $$n \geq \frac{2010 \cdot 66}{42 \cdot 67}=\frac{330}{7}>47$$ 所以 $n \geq 48$。现在我们需要证明 $n=48$ 适合。考虑集合 $\{2,3, \ldots, 33,36,37, \ldots, 50,67\}$，它恰好包含 48 个数字。我们有 $$\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{32}{33} \cdot \frac{35}{36} \cdots \frac{49}{50} \cdot \frac{66}{67}=\frac{1}{33} \cdot \frac{35}{50} \cdot \frac{66}{67}=\frac{7}{335}=\frac{42}{2010}$$ 因此对于 $n=48$ 存在一个所需的示例。评论 1. 在这个版本的问题中，估计需要多一步，因此有点困难。另一方面，这个版本中的例子并不是唯一的。另一个例子是 $$\frac{1}{2} \cdot \frac{2}{3} \cdots \frac{46}{47} \cdot \frac{66}{67} \cdot \frac{329}{330}=\frac{1}{67} \cdot \frac{66}{330} \cdot \frac{329}{47}=\frac{7}{67 \cdot 5}=\frac{42}{2010} .$$ Comment 2. N1' 是提案者对问题的表述。我们根据当前IMO的数量提出N1。