10.1. Find all pairs of natural numbers $x \geq 2, y \geq 2$ such that the remainder of the division of the number $3 \boldsymbol{x}$ by $\boldsymbol{y}$ is 1, the remainder of the division of the number $3 \boldsymbol{y}$ by $\boldsymbol{x}$ is 1, and the remainder of the division of the number $\boldsymbol{x} \boldsymbol{y}$ by 3 is also 1.
Answer. Four solutions: $x=2, y=5, x=5, y=2$.

Solution. The numbers $x$ and $y$ cannot be equal, otherwise the remainder of the division of the number $3x$ by $y$ would be 0. Therefore, due to symmetry, we can assume $x<y$, so $3x-1<3y$ and $3x-1$ is divisible by $y$, from which $3x-1=y$ or $3x-1=2y$. In the first case, $3y-1=9x-4$ is divisible by $x$, meaning 4 is divisible by $x$, hence $x=2, y=5$ or $x=4, y=11$. In the first case, the remainder of the division of the number $xy$ by 3 is 1, so it satisfies the condition of the problem. In the second case, the remainder of the division of the number $xy$ by 3 is 2, so it does not satisfy the condition of the problem.

In the second case, $2(3y-1)=6y-2=9x-5$ is divisible by $x$, meaning 5 is divisible by $x$, hence $x=5, y=7$. But in this case, the remainder of the division of the number $xy$ by 3 is 2, and the pair $x=5, y=7$ does not satisfy the condition.

Grading criteria. ($\cdot$) Correct solutions found by trial: 1 point for $x=2, y=5$, and 1 point for $x=5, y=2$: a total of 2 points. ($\cdot$) Noted that $3x-1=y$ or $3x-1=2y$: 2 points. ($\cdot$) Correctly analyzed the first of these cases: 3 points. Correctly analyzed the second of these cases: 2 points.