
Problem 3. Find the smallest positive integer $n$, for which there exist $n$ different positive integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying the conditions:

a) the smallest common multiple of $a_{1}, a_{2}, \ldots, a_{n}$ is 1995 ;

b) for each $i, j \in\{1,2, \ldots, n\}$ the numbers $a_{i}$ and $a_{j}$ have a common divisor $\neq 1$;

c) the product $a_{1} a_{2} \ldots a_{n}$ is a perfect square and is divisible by 243 .

Find all $n$-ples $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$, satisfying a), b) and c).


Solution. Since $1995=3.5 .7 .19$ and $a_{i} / 1995$, then for all $i$

$$
a_{i}=3^{\alpha_{i}} 5^{\beta_{i}} \gamma^{\gamma_{i}} 19^{\delta_{i}}
$$

where the numbers $\alpha_{i}, \beta_{i}, \gamma_{i}, \delta_{i}$ are equal to 0 or 1 . There is no $a_{i}$ which is divisible by 9 . We have $a_{1} a_{2} \ldots a_{n}=k^{2}$, where $k$ is a positive integer and since $243=3^{5}$ divides $k^{2}$, then $3^{6}$ divides $k^{2}$ and since among the numbers (*) there is no one which is divisible by 9 , then $n \geq 6$. The numbers (*), which are divisible by 3 are

$$
3 ; \quad 3.5 ; \quad 3.7 ; \quad 3.19 ; \quad 3.5 .7 ; \quad 3.5 .19 ; \quad 3.7 .19 ; \quad 3.5 .7 .19
$$

They are 8 in number. Let $n=6$. Then the numbers $a_{1}, a_{2}, \ldots, a_{n}$ are among (**). It is easy to see that the product of any 6 numbers from $(* *)$ is not a perfect square. Thus $n \geq 7$. Let $n=7$. It is not possible that all the numbers $a_{1}, a_{2}, \ldots, a_{n}$ are divisible by 3 , because in such a case $3^{7} / k^{2}$ and $3^{8}+k^{2}$. Therefore 6 numbers from $a_{1}, a_{2}, \ldots, a_{n}$ are divisible by 3 , i.e. they are from $(* *)$ and at least one of them (for example $a_{1}$ ) is not divisible by 3 . It follows from a) that among $a_{1}, a_{2}, \ldots, a_{7}$ there is at least one which is divisible by 5 , at least one which is divisible by 7 and at least one which is divisible by 19. Since each pair of these numbers has a common divisor, and $3+a_{1}$, then $a_{1}$ must be divisible by 5.7.19, i.e. $a_{1}=5.7 .19$. At
last note that the product of all numbers $(* *)$ is equal to $3^{8} \cdot 5^{4} \cdot 7^{4} \cdot 19^{4}$. The only possibility is $a_{1} a_{2} \ldots a_{7}=3^{6} \cdot 5^{4} \cdot 7^{4} \cdot 19^{4}$. From here $a_{2} a_{3} \ldots a_{7}=3^{6} \cdot 5^{3} \cdot 7^{3} \cdot 19^{3}$ and the only possibility is

$$
\left\{a_{2}, a_{3}, \ldots, a_{7}\right\}=\{3 \cdot 5,3 \cdot 7,3 \cdot 19,3 \cdot 5 \cdot 7,3 \cdot 5 \cdot 19,3 \cdot 7 \cdot 19\}
$$

Therefore $n=7$ and

$$
\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}\right\}=\{3 \cdot 5,3 \cdot 7,3 \cdot 19,3 \cdot 5 \cdot 7,3 \cdot 5 \cdot 19,3 \cdot 7 \cdot 195 \cdot 7 \cdot 19\}
$$

## Grade 11.
