ISL 2010 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.