\[{ ( 1-\frac{1}{s_{1}})( 1-\frac{1}{s_{2}})\cdots( 1-\frac{1}{s_{n}}) \leq ( 1-\frac{1}{2})( 1-\frac{1}{3}})\cdots( 1-\frac{1}{n+1})=\frac{1}{n} \], so $n\geq 39$. Now \[ ( 1-\frac{1}{2})( 1-\frac{1}{3})\cdots( 1-\frac{1}{33})\times ( 1-\frac{1}{35})( 1-\frac{1}{36})\cdots( 1-\frac{1}{40}) \times ( 1-\frac{1}{67})=\frac{1}{33} \times \frac{34}{40} \times \frac{66}{67}=\frac{51}{2010}\]. Therefore $n=39$.

In fact that inequality is reversed (and it should be mentioned that if any of the $s_k$'s is $1$, the expression is null, so we need all $s_k \geq 2$) \[{ ( 1-\frac{1}{s_{1}})( 1-\frac{1}{s_{2}})\cdots( 1-\frac{1}{s_{n}}) \geq ( 1-\frac{1}{2})( 1-\frac{1}{3}})\cdots( 1-\frac{1}{n+1})=\frac{1}{n}. \] But $\frac {1} {40} < \frac {51}{2010} < \frac {1} {39}$, so $n\geq 39$.

The shortlist also contained a similar problem N1', in which the number 51 was replaced by 42. According to a comment on the shortlist, the original formulation of the problem used 42, but the PSC changed this into 51 because it was the 51st IMO. The solution to this problem is also similar:

What about this "similar" problem: Find the least positive integer $n$ for which there exists a set $\{s_1, s_2, \ldots , s_n\}$ consisting of $n$ (not necessarily distinct) positive integers such that \[ \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) = \frac{51}{2010}.\]

Actually it is much easy to show that $n \ge 39$,and the real task is to find out the $s_i$'s for $n=39$. $\frac{51}{2010}$ is nothing but $\frac{17}{670}$ so one of the $s_i$'s should be $67$(or its multiple).Also one of the $s_i-1$'s is divisible by $17$ so it is reasonable to think of one being $34$.Also the numerator of $1-\frac{1}{67}$ is $66$ so one of the $s_i$'s should be $33$.That is all about how to guess and come up with the solution $\{2,3,\dots,33,35,36,37,38,39,40,67\}$

First, we do some bounding. Since all the $s_j$ are distinct, $s_j \ge j+1$, so $1 - \frac{1}{s_j} \ge \frac{j}{j+1}$. Therefore, \[ \frac{51}{2010} \ge \frac{1}{n+1} \implies n \ge 39 \]It suffices to show that there exists a valid set with $39$ elements. Here's a solution with motivation. Since $\frac{51}{2010} = \frac{17}{2 \cdot 5 \cdot 67}$, we want a $67$. Also, we need a $17$ and need to cancel $66$, so we take $18, 19, \cdots, 33, 67$. Then to get $2 \cdot 5$ on the denominator we take $2, 3, \cdots, 20$, since all other obvious options yield over $39$ elements, but this doesn't work either. Thus, we get a $17$ indirectly instead, with a $34$. Again we want a $33$ so take $2, \cdots, 33, 35, 67$. It turns out that we can finish with $36, \cdots, 40$. So educated guessing tells us that the set $2, \cdots, 33, 35, 36, 37, 38, 39, 40, 67$ works and we are done.

Amir Hossein wrote: Find the least positive integer $n$ for which there exists a set $\{s_1, s_2, \ldots , s_n\}$ consisting of $n$ distinct positive integers such that \[ \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) = \frac{51}{2010}.\] Proposed by Daniel Brown, Canada

The answer is $n = 39$. To see this is optimal, assume $s_i > 1$ for all $i$ forever after. Then for any $n$, \[ \prod \left(1 - \frac{1}{s_i}\right) \ge \left(1-\frac12\right) \left(1-\frac13\right) \dots \left(1-\frac1{n+1}\right) = \frac{1}{n+1} \]and since $\frac{51}{2010} < \frac{1}{39}$, we need $n+1 > 39$, or $n > 38$. As for a construction when $n = 39$, note that \[ \left( \frac12 \cdot \frac23 \cdot \dots \cdot \frac{32}{33} \right) \cdot \left( \frac{34}{35} \cdot \dots \cdot \frac{39}{40} \right) \cdot \frac{66}{67} \]works, since it equals $\frac{1/40}{33/34} \cdot \frac{66}{67} = \frac{17}{670} = \frac{51}{2010}$.

SL 2010 has such short problems! Amir Hossein wrote: Find the least positive integer $n$ for which there exists a set $\{s_1, s_2, \ldots , s_n\}$ consisting of $n$ distinct positive integers such that \[ \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) = \frac{51}{2010}.\] Proposed by Daniel Brown, Canada Answer : $n=39$ $ $ Firstly, none of $s_j \neq 1 \forall 1 \le j \le n$. Then proceed with, $$\prod_{cyc} \left(1-\frac{1}{s_1}\right) \ge \left(1-\frac{1}{2}\right) \cdot \dots \left(1-\frac{1}{n+1}\right) = \frac{1}{n+1}$$ Since, $\frac{51}{2010} < \frac{1}{39} \iff n \ge 39$. Now as a construction, we have the set $\left(\frac{1}{2} , \frac{2}{3} , \dots , \frac{32}{33} , \frac{34}{35} , \dots , \frac{39}{40} , \frac{66}{67}\right)$

The answer is $n = 39$. Note that \[ \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) \le \frac{1}{2} \cdot \frac{2}{3} \cdots \frac{n}{n+1} = \frac{1}{n}\]so $n \ge \lfloor \frac{2010}{51} \rfloor = 39.$ For a construction, take $$(s_1, s_2, \dots s_39) = (2, 3, \dots 33, 35, \dots 40, 67),$$and it's easy to see that this works. $\blacksquare$

It is clear that $s_i\ge 2$ for each $i$. Remark that \[\left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) \ge \frac{1}{n+1}.\]Hence, we must have \[n \ge \left\lceil \frac{2010}{51}\right\rceil-1 = 39.\]Now, note that the following construction works \[\{s_1,\dots , s_{39}\} = \{2,3,\dots , 33,35,36,\dots , 40, 67\},\]as it yields \[\left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) = \frac{1}{33} \cdot \frac{34}{40} \cdot \frac{66}{67} = 2 \cdot \frac{17}{20} \cdot \frac{1}{67} = \frac{17}{670} = \frac{51}{2010}.\] Remarks: The hard part is guessing that there has to be a construction. After that, taking $s_{39}=67$ and manipulating a bit allows one to find this construction.

Took me 45 minutes to find construction and 5 minutes to prove lower bound The answer is $n = 39$, with the following construction: \[s_{1} = 2, s_{2} = 3, \ldots s_{32} = 33, s_{33} = 35, s_{34} = 36, \ldots s_{38} = 40, s_{39} = 67\]This construction results in $\frac{1}{33} \cdot \frac{34}{40} \cdot \frac{66}{67} = \frac{17}{670} = \frac{51}{2010}$. We can not do better, since if $n < 39$, then the minimal possible value is $\frac{1}{39} > \frac{51}{2010}$

First remark the "dumb" bound: $$\left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) \geq \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{3} \right) \cdots \left( 1 - \frac{1}{n+1} \right)$$hence $n+1 \geq \tfrac{2010}{51}$ and thus $n \geq 39$. It remains to display a construction for $n=39$. Indeed, take $$(s_i) = \{ 2,3,\dots,33 \} \quad \cup \quad \{ 35,36,\dots,40 \} \quad \cup \quad \{ 67\}. $$ This construction works; note that $$\left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) = \left(\frac{1}{33}\right) \left(\frac{34}{40} \right) \left(\frac{66}{67} \right) = \frac{51}{2010},$$as desired.

We note that the product when we have $n$ numbers is at least \[\prod_{i=1}^{n} \left(1-\frac{1}{i}\right) = \frac{1}{n+1}\]Thus, we must have $\frac{51}{2010}>\frac{1}{n+1}\Longrightarrow n\geq 39$ We now construct in the $n=39$ case. We take the set of $s_i$ of $[1,39]+[67]-[34]$ which gives a product of \[\frac{1}{2}\cdot \frac{2}{3}\cdots \frac{32}{33}\cdot \frac{34}{35}\cdot \frac{35}{36}\cdots \frac{39}{40}\cdot \frac{66}{67}=\frac{1}{33}\cdot \frac{34}{40}\cdot \frac{66}{67}=\frac{51}{2010}\]And we have shown that $n=39$ is both the minimum and achievable, so we are done $\blacksquare$.

ISL 10N1. Find the least $n\in\mathbb{Z}$ such that $\exists S=\{s_i\in\mathbb{Z}_+: 1\le i\le n, i\in\mathbb{Z}\}$, where $|S|=n$ and \[\prod_{s_i\in S}\left(1-\frac{1}{s_i}\right)=\frac{51}{2010}.\] Solution. Note that \[\frac{51}{2020}=\prod_{s_i\in S}\left(1-\frac{1}{s_i}\right)\ge\left(1-\frac12\right)\left(1-\frac13\right)\ldots\left(1-\frac1{n+1}\right)=\frac1{n+1}\implies n\ge 39.\]\[\left(\frac12\cdot\frac23\cdot\ldots\cdot\frac{32}{33}\right)\cdot\left(\frac{34}{35}\cdot\ldots\cdot\frac{39}{40}\right)\cdot\frac{66}{67}=\frac{51}{2010}.\]Therefore the least such $n\in\mathbb{Z}$ is $39$.

The answer is $\boxed{n=39}$. We first show a construction, then show this is optimal. We have $\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}\cdot \cdots \cdot \frac{32}{33}\cdot \frac{34}{35}\cdot \frac{35}{36}\cdot \cdots \cdot \frac{39}{40}\cdot \frac{66}{67}$. Now, notice that in order to decrease the value of the LHS as fast as possible, we should try to make the $s_i$ as large as possible. Notice that to the maximize $n=38,$ the smallest value we could have for the LHS is $\frac{1}{39}$(just by letting $s_k=k$ for $1\leq k\leq 38$). It is easy to see that $\frac{1}{39}\geq \frac{17}{2\cdot 5\cdot 67},$ so we are done.

We claim that the answer is $\boxed{n=39}$. This can be achieved with the set $\{s_1, s_2, \ldots , s_n\}$ being equal to $\{2,3,\ldots, 33,35,\ldots, 39, 40, 67\}$, essentially every number from $2$ to $40$ except $34$ in addition to $67$. It is easy to see that this works. We will now prove that it is necessary that $n\ge 39$. Notice that none of the $s_i$ can be $1$, and the function $f(x)=1-\frac{1}{x}$ is increasing, so to minimize $$\left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right),$$we should have $\{s_1, s_2, \ldots , s_n\}=\{2,3,\ldots,n+1\}.$ However, it is easy to see that using this set yields $\frac{1}{n+1}$, therefore $$\frac{51}{2010}\ge \frac{1}{n+1} \Rightarrow n\ge 39,$$as desired. Remarks: This is rather easy for the IMO SL, even at the first spot, it was literally a ten-minute solve. Also, this is the first ISL problem in which I have seen an unsimplified fraction.

We claim the answer is $n=39.$ Notice that $$\frac{17}{670}=\prod_{i=1}^{n}{\left(1-\frac{1}{s_i}\right)}\ge\prod_{i=2}^{n+1}{\left(1-\frac{1}{i}\right)}=\frac{1}{n+1}.$$Hence, $n\ge670/17-1>38$ so $n\ge39.$ Also, $n=39$ works as $$\left(1-\frac{1}{67}\right)\prod_{i=2}^{33}{\left(1-\frac{1}{i}\right)}\prod_{i=35}^{40}{\left(1-\frac{1}{i}\right)}=\frac{66}{67}\cdot\frac{1}{33}\cdot\frac{34}{40}=\frac{17}{670}.$$$\square$

We claim that the answer is 39. Clearly, it is not possible for $n\leq38$ since the smallest value we can achieve with only 38 terms is $$(1-1/2)(1-1/3)\cdots (1-1/39)=1/39>17/660,$$so it is not possible. We now show that 39 terms is possible. This is achieved by $$(s_1,s_2\cdots s_{39})=(2,3,4,5\cdots32,33,35,36,37,38,39,40,67).$$Essentially, we take 2 through 40, remove 34, and add back 67. This clearly works as $$\frac{1}{33}\cdot \frac{34}{40}\cdot \frac{66}{67}=\frac{17}{670}.$$ Remark: The motivation for the construction was the following. We obviously want a $66/67$ factor since we need a 67 in the denominator. Then, we need the remaining product to be $17/660$ with 38 terms. This is barely larger than $1/39$, so we need to stick pretty close to 2 through 39. Additionally, we need to "skip" a multiple of 17 in order to get the 17 in the numerator (so that it doesn't cancel away), and the construction is pretty easy to discover from there.

We claim that there must be at least $39$ numbers. First, we must prove that $39$ integers works, which can be done by doing \[\frac{1}{2}\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\dots{}\left(\frac{1}{33}\right)\left(\frac{34}{35}\right)\left(\frac{35}{36}\right)\dots{}\left(\frac{39}{40}\right)\left(\frac{66}{67}\right)=\frac{51}{2010}.\] Since the integers are all distinct, we must have that \[\left(1-\frac{1}{s_1}\right)\left(1-\frac{1}{s_2}\right)\dots{}\left(1-\frac{1}{s_n}\right)\geq{}\frac{1}{n+1}.\]However, since $\frac{1}{40}<\frac{51}{2010}<\frac{1}{39}$, $n$ must be at least $39$, and we are done. Problem Variation (My inability to read the full problem statement): Find the least positive integer $n$ for which there exists a set $\{s_1, s_2, \ldots, s_n\}$ consisting of $n$ (not necessarily distinct) positive integers satisfying \[ \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \dots \left( 1 - \frac{1}{s_n} \right) = \frac{51}{2010}.\] We claim that we must have at least $8$. This can be achieved through \[\left(\frac{1}{2}\right)^3\left(\frac{4}{5}\right)^2\left(\frac{30}{31}\right)\left(\frac{66}{67}\right)\left(\frac{527}{528}\right).\] Now we must prove that we must have at least $8$. Notice that we must have at least one fraction with a multiple of $17$ in the numerator, one with $67$ in the denominator, and one with $5$ in the denominator. Taking casework, we find that the minimum number of numbers used to get the final number under $\frac{51}{2010}$ occurs when we use $\left(\frac{17}{18}\right)\left(\frac{66}{67}\right)\left(\frac{4}{4}\right)\left(\frac{1}{2}\right)^5$, which requires at least $8$ numbers, and we are done.

This proof is composed of 2 parts. Part 1: proof of minimality We claim that 39 is the smallest number. To prove this, we show that any collection of 38 fractions will result in something greater. The smallest collection is obviously the first 38 such fractions (these fractions increase as s(i) increase. By telescoping they multiply to 1/39, which is greater than 51/2010. So therefore 38 does not work. However, 39’s minimality is 1/40 which is less than 51/2010, so 39 is potentially a solution. Part 2: existence of a collection Try to construct such a set with brute force (a shortcut is to realize that it must have 66/67). The collection for s(i) are 2, 3,……33, 35, 36, 37, 38, 39, 40, 67. The answer to this question is then N = 39

Clearly $n \geq 39$ since \[\left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) \geq \left(1-\frac12\right) \left(1-\frac13\right) \dots \left(1-\frac1{n+1}\right) = \frac{1}{n+1}\]so we must have $\frac{51}{2010} \geq \frac{1}{n+1} \implies n \geq 39$. A construction for $n = 39$ is \[(s_1,s_2\cdots s_{39})=(2,3,4,5\cdots32,33,35,36,37,38,39,40,67),\]which clearly works as this product is equal to \[\frac{1}{33}\cdot \frac{34}{40}\cdot \frac{66}{67}=\frac{17}{670} = \frac{51}{2010}.\]

Notice that $$\frac{51}{2010}=\left(1-\frac1{s_1}\right)\left(1-\frac1{s_2}\right)\cdots\left(1-\frac1{s_n}\right) \ge \left(1-\frac12\right)\left(1-\frac13\right)\cdots\left(1-\frac1{n}\right)=\frac1{n+1},$$and solving the inequality gives $n \ge 39$. As a construction, note that $$\left(1-\frac12\right)\left(1-\frac13\right)\cdots\left(1-\frac1{33}\right)\left(1-\frac1{35}\right)\left(1-\frac1{36}\right)\cdots\left(1-\frac1{40}\right)\left(1-\frac1{67}\right)=\left(\frac1{33}\right)\left(\frac{34}{40}\right)\left(\frac{66}{67}\right)=\frac{51}{2010},$$which works. $\blacksquare$

same sol as literally everyone else since this one was easy, but whatever.

Note that $s_i\ge i+1$, so \[\left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \dots \left( 1 - \frac{1}{s_n} \right)\ge\frac12\cdot\frac23\cdot\frac34\cdot\ldots\frac{n}{n+1}=\frac{1}{n+1}\]As $\frac{51}{2010}<\frac{1}{39}$, we have $n+1>39\implies n\ge 39$. Now the hard part is indeed the construction. Let $s_{39}=66$ so that we have the factor $\frac{66}{67}$. Now, we can cancel the $66$ partly with $\frac12\cdot\frac23\cdot\ldots\cdot\frac{32}{33}=\frac{1}{33}$. This gets us the term $\frac{2}{67}$. It remains to construct the term $\frac{17}{4\cdot 5}=\frac{34}{40}$, which is easy. Now, \[\{s_1 < s_2 < \ldots < s_n\}=\{1,2,3,\ldots,33,35,36,\ldots,40,63\}\]is a solution which contains $39$ terms.

The answer is $n = 39$. Note that $n \geq 39$ because $$\prod_{i=1}^n \left(1-\frac 1{s_i}\right) \leq \frac 1{n+1}.$$On the other hand $\{s_1, s_2, \dots, s_n\}=\{1, 2, \dots, 33, 35, 36, \dots, 40, 63\}$ works.

We claim that $n=39$. Assume WLOG that $1\leq s_1 < s_2 < \cdots < s_n$. So, $s_i\geq i$. Hence, $1-\frac1{s_i}\geq 1-\frac1i = \frac{i-1}i$. Hence, we have \[\frac{51}{2010} = \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) \geq \frac12 \cdot \frac23 \cdots \frac{n-1}n = \frac 1n \Longrightarrow n\geq 39.\] This proves the bound. We now give a construction. Note that $\{s_1,s_2,\ldots,s_n\} = \{1,2,\ldots,33,35,36,\ldots,40,63\}$ works.

Sketch: $n \geq 39$ since $\frac{1}{n+1} \leq \frac{17}{670}$. Now for the construction just use that $\frac{17}{670}=\frac{66 \times 34 \times 1}{67 \times 40 \times 33}$

I claim the answer is just $n = 39$, we can get the bound by size since if $n < 39 $ the product must be at least $\frac 12 \frac 23 \cdots \frac{38}{39} = \frac{1}{39} > \frac{17}{670} $, and for the construction take $2, \cdots 33, 35, \cdots 40, 67 = \frac{1}{33} \frac{17}{20} \frac{66}{67} = \frac{17}{670}$

We have that $\frac{1}{n}\leq \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \dots \left( 1 - \frac{1}{s_n} \right)$ which is clearly true by substituting in $2, 3, \dots, n+1$, thus as a construction exists for $39$ this suffices.

The answer is $39.$ Notice that, if $n \le 38,$ then $$\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \dots \left(1-\frac{1}{s_{n}}\right) > \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{38}{39} > \frac{51}{2010}.$$Now, if the set is $\{1, 2, \dots, 33\} \cup \{35, \dots, 40\} \cup \{67\},$ we get $$\frac{1}{33} \cdot \frac{34}{40} \cdot \frac{66}{67} = \frac{51}{2010}, $$as desired.

Note that $$\frac{51}{2010} = \left( 1 - \frac{1}{s_1} \right) \left( 1 - \frac{1}{s_2} \right) \cdots \left( 1 - \frac{1}{s_n} \right) \geq \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{3} \right) \cdots \left( 1 - \frac{1}{n+1} \right) = \frac{1}{n+1} \implies n \geq 39.$$Now, to show that this works consider $$\left( \frac{1}{2} \cdot \frac23 \cdots \cdots \cdot \frac{32}{33} \right) \cdot \left( \frac{34}{35} \cdot \frac{35}{36} \cdot \cdots \cdot \frac{39}{40} \right) \cdot \left( \frac{66}{67} \right) = \frac{17}{670} = \frac{51}{2010}.$$Therefore $(s_1, s_2, \cdots, s_{39}) = (2, 3, 4, \cdots, 33; 35, 36, \cdots 40; 67)$ works, so the minimum is $n=39.$

Please contact westskigamer@gmail.com if there is an error with my solution for cash bounties by 3/18/2025.