An olympiad adventure

This is problem N1 (proposed by Canada) of the 51st IMO Shortlist (2010).

Official solution:

Suppose that for some $p$ there exist the desired numbers; we may assume that $\lambda_1<\lambda_2<\ldots<\lambda_p$ .

Surely $\lambda_1>1$ since otherwise $1-\dfrac{1}{\lambda_1}=0$ .

So we have $2\le\lambda_1\le\lambda_2-1\le\ldots\le\lambda_p-(p-1)$ , hence $\lambda_i\ge i+1$ for each $i=1, \ldots, p$ .

Therefore,

$\dfrac{51}{2010}=\left( 1-\dfrac{1}{\lambda_1} \right) \left( 1-\dfrac{1}{\lambda_2} \right) \ldots \left( 1-\dfrac{1}{\lambda_p}\right)$

$\qquad\;\;\ge \left( 1-\dfrac{1}{2} \right) \left( 1-\dfrac{1}{3} \right) \ldots \left( 1-\dfrac{1}{p+1}\right)$

$\qquad\;\;=\dfrac{1}{2}.\dfrac{2}{3}.\ldots\dfrac{p}{p+1}=\dfrac{1}{p+1}$ ,

which implies $p+1\ge\dfrac{2010}{51}>39$ , so $p\ge39$ .

Now we are left to show that $p=39$ fits. Consider the set $\{2, 3, \ldots, 33, 35, 36, \ldots, 40, 67\}$ which contains exactly $39$ numbers. We have:

$\dfrac{1}{2}.\dfrac{2}{3}.\ldots\dfrac{32}{33}.\dfrac{34}{35}.\ldots\dfrac{39}{40}.\dfrac{66}{67}=\dfrac{1}{33}.\dfrac{34}{40}.\dfrac{66}{67}=\dfrac{17}{670}=\dfrac{51}{2010}$

hence for $p=39$ there exists a desired example.