Khachatryan M.A.

On the front side of each of the $\$ 6 \$$ cards, Anya wrote a natural number with a black or red marker. At the same time, Anya wrote at least two numbers with each color.

Then Borya took each card, looked at the color of the number on it, multiplied all Anya's numbers of the same color on other cards, and wrote the result on the back of the card (if there was only one other card of the same color, Borya wrote the number from that one card).

We see the backs, on which the numbers $\$ 18 \$, \$ 23 \$, \$ 42 \$, \$ 42 \$, \$ 47 \$ 2$, What are the numbers written on the front sides of these cards?
Let's look at a card with the inscription “\$23\$”. Suppose the number on the front side of this card is written in red. Then among the red numbers, there is a number that is divisible by \$23\$. If there are at least three red numbers, on the backs there should be at least two numbers that are divisible by \$23\$, but this is not the case. Therefore, there are exactly two red numbers, and one of them is on the front side of the card with the number \$23\$. The four numbers written in black are then written in black.

Similar reasoning can be applied to the number \$47\$: the color of the number on the front side of the card with the number \$47\$ appears twice, i.e., it is red. But then on the front side of the card with the number \$23\$ is written the number \$47\$, and on the card with the number \$47\$ is written the number \$23\$.

Now let's consider the cards on which Anya wrote the numbers in black. On their backs, the numbers \$18\$, \$42\$, \$42\$, and \$63\$ are written. In their product \$18 \cdot 42 \cdot 42 \cdot 63 = 126^3\$, each of the black numbers appears three times. Therefore, the product of the black numbers is \$126\$. Then on the front side of the card with the number \$18\$ is written the number \$126 / 18 = 7\$; with the number \$42\$ is written the number \$126 / 42 = 3\$; with the number \$63\$ is written the number \$126 / 63 = 2\$.

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_53798a1e86d520071242g-29.jpg?height=52&width=867&top_left_y=2784&top_left_x=1)

![](https://cdn.mathpix.com/cropped/2024_05_06_53798a1e86d520071242g-30.jpg?height=45&width=263&top_left_y=129&top_left_x=0)

Four natural numbers are given. Each of these numbers is divisible by the greatest common divisor of the other three. The least common multiple of any three of these numbers is divisible by the remaining fourth. Prove that the product of these numbers is a perfect square.

## Solution

Let the given numbers be \$\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\$. Let \$\mathrm{p}\$ be an arbitrary prime number. Denote the exponents of \$\mathrm{p}\$ in these numbers by \$\alpha, \beta, \gamma, \delta\$ respectively. We can assume that \$\alpha \geqslant \beta \geqslant \gamma \geqslant \delta\$.

Since \$d\$ is the smallest, \$\delta\$ is the smallest exponent. Therefore, \$\delta \leqslant \alpha - 1\$.

Since \$[\mathrm{~b}, \mathrm{c}, \mathrm{d}]\$ is divisible by \$a\$,

Therefore, \$p\$ enters \$abcd\$ in an even power \$2\alpha + 2\delta\$. Since this is true for any \$\mathrm{p}\$,

Thus, the product of the given numbers is a perfect square.