Find the least positive integer $n$ so that $\left ( 1-\frac{1}{s_{1}} \right ) \cdots \left ( 1-\frac{1}{s_{n}} \right )=\frac{51}{2010}$ 
Find the least positive integer $n$  for which there exists a set $\left \{ s_{1}, s_{2},....,s_{n} \right \}$ consisting of $n$ distinct positive integers such that $$\left ( 1-\frac{1}{s_{1}} \right ) \left ( 1-\frac{1}{s_{2}} \right )\cdots\left ( 1-\frac{1}{s_{n}} \right )=\frac{51}{2010}$$

My idea was assume that $s_{1}< s_{2}< \cdots< s_{n}$, So I can't know how to start, any help will be appreciate it.
 A: Suppose that for some $n$ there exist the desired numbers; we may assume that $s_{1}< s_{2}< ...< s_{n}$. Surely $s_{1}> 1$ since otherwise $1-\frac{1}{s_{1}}=0 $. So we have $2\leq s_{1}\leq s_{2}-1\leq s_3 - 2\leq ....\leq s_{n} - (n-1)$, hence $s_{i}\geq i+1$ for each $i=1,...,n$. Therefore $$\frac{51}{2010}=\left ( 1-\frac{1}{s_{1}} \right )\left ( 1-\frac{1}{s_{2}} \right )...\left ( 1-\frac{1}{s_{n}} \right )\geq \left ( 1-\frac{1}{2} \right )\left ( 1-\frac{1}{3} \right )....\left ( 1-\frac{1}{n+1} \right )=\frac{1}{2}\cdot \frac{2}{3} ... \frac{n}{n+1}=\frac{1}{n+1}$$
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 $\left \{ 2,3,...,33,35,36...,40,67 \right \}$ which contains exactly $39$ numbers.   We have $$\frac{1}{2}\cdot \frac{2}{3}....\frac{32}{33}\cdot \frac{34}{35}...\frac{39}{40}\cdot \frac{66}{67}=\frac{17}{670}=\frac{51}{2010}$$
Hence for $n=39$ there exists a desired example.
AND WE'RE DONE
