Estonian Mathematical Olympiad

Let's first count the choices that include the number $1$. If some three of the remaining numbers were $2$, $3$, and $4$, then the conditions of the problem would be violated, because the denominators of all the fractions in the list are smaller than $2\cdot 3\cdot 4=24$, and the product of the selected 5 numbers could not be $1$. If instead of $2$, $3$, and $4$ some other 3 integers are chosen, their product will be even larger, so the conditions of the problem cannot be met. The situation is similar if 3 fractions are chosen next to the number $1$. Thus, in addition to the number $1$, the choice must include 2 integers and 2 fractions, with the product of the integers equal to the product of the denominators of the fractions. There are exactly 21 ways to choose 2 integers in addition to the number $1$, and the resulting products are in the following table:

	3	4	5	6	7	8
2	6	8	10	12	14	16
3		12	15	18	21	24
4			20	24	28	32
5				30	35	40
6					42	48
7						56

We see that the numbers $12$ and $24$ appear in the table 2 times, the remaining numbers are unique. Thus, we get 21 options where the selected integers are the same as the denominators of the selected fractions, and in addition 4 options where the selected integers and the denominators of the selected fractions are not the same, but their product is either $12$ or $24$. In total, there are 25 options with the number $1$.

Now we count the options that do not have the number $1$. We have to choose either 3 integers and 2 fractions or 3 fractions and 2 integers. Since, due to symmetry, there are the same number of options of both types, we count the options involving 3 integers and 2 fractions. If $2$ were not included in the selection, the product of the selected integers should be at least $3\cdot 4\cdot 5=60$, but according to the multiplication table, the product of the denominators of any 2 fractions is less than $60$, which is why the condition of the problem cannot be met. Therefore, $2$ must be included in the selection. Similarly, we see that either $3$ or $4$ must also be included in the selection. We get the possibilities $$\begin{gathered} 2\cdot 3\cdot 4=24,2\cdot 3\cdot 5=30,2\cdot 3\cdot 6=36,2\cdot 3\cdot 7=42,2\cdot 3\cdot 8=48, \\ 2\cdot 4\cdot 5=40,2\cdot 4\cdot 6=48,2\cdot 4\cdot 7=56,2\cdot 4\cdot 8=64. \end{gathered}$$ Of these 9 products, only $36$ and $64$ do not appear in the multiplication table above, the product $24$ appears there 2 times and the remaining 6 products 1 time. So we get $2\cdot 0+1\cdot 2+6\cdot 1=8$ possibilities. There are the same number of possibilities with 3 fractions and 2 integers. So there are $2\cdot 8=16$ possibilities without the number $1$. Hence the total number of possibilities is $25+16=41$.

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Alternative solution.

The list of numbers contains 7 pairs of reciprocals in the form $(\frac{1}{a},a)$. At most 2 such pairs can occur among the 5 numbers chosen. If there are two pairs of reciprocals in the choice, then the last number must be $1$. To choose 2 pairs from the 7 existing pairs of reciprocals, there are $\frac{7\cdot 6}{2}=21$ possibilities. If there is one pair of reciprocals in the choice, then the product of the remaining 3 numbers must be $1$. If the integers $a$ and $b$ are among these 3 numbers, then the last number must be $\frac{1}{ab}$, so the integer $ab$ is also in the original list. Similarly if the fractions $\frac{1}{a}$ and $\frac{1}{b}$ are among these 3 numbers, then $a$, $b$ and $ab$ must be in the original list. The only such numbers are $2$, $3$, $6$ and $2$, $4$, $8$, each of which gives 2 choices for the 3 numbers from the original list. Then there are $7-3=4$ options for the pair of reciprocal numbers. So in this case there are $2\cdot 2\cdot 4=16$ possibilities. If there are no pairs of reciprocals in the selection, then each integer can appear in the selection at most once, either by itself or in the denominator of the reciprocal. The prime numbers $5$ and $7$ cannot appear in either role. The prime number $3$ only appears as a factor in the numbers $3$ and $6$; if both of them were omitted, there would not be enough numbers left in the selection, so they must both be in the selection, either as $\frac{1}{3}$ and $\frac{1}{6}$ or as $\frac{2}{9}$ and $\frac{2}{3}$. The product of these numbers is $\frac{1}{2}$ or $2$, respectively. The remaining numbers are $1$, $2$, $4$, and $8$, from which 3 must be chosen. If one does not choose $1$, then it is not possible to get the number $2$ or $\frac{1}{2}$ by using each remaining number or its reciprocal as a factor only once. Thus, the number $1$ must be in the selection. From the remaining numbers and their reciprocals, the number $2$ or $\frac{1}{2}$ can be composed in 4 ways: $\frac{1}{2}\cdot 4$, $\frac{1}{4}\cdot 8$, $2\cdot \frac{1}{4}$, and $4\cdot \frac{1}{8}$. So the total number of choices is $21+16+4=41$.