10.2. Find all pairs of natural numbers $x$ and $y$ such that their least common multiple is equal to $1 + 2x + 3y$.
Answer: $x=4, y=9$ or $x=10, y=3$.

Solution. Let's first assume $x \leq y$. Note that $y$ cannot be divisible by $x$, otherwise the least common multiple of $x$ and $y$ would be $y$, which is less than $1+2x+3y$. In particular, $x>1$.

Next, the least common multiple of $x$ and $y$ is divisible by $x$ and $y$, so $1+2x+3y$ is divisible by $x$ and $y$, which means $1+2x$ is divisible by $y$ and $1+3y$ is divisible by $x$. From the divisibility of $1+2x$ by $y$, it follows that $1+2x=ky \geq y$, which together with the assumption $x \leq y$ implies $k=1, y=2x+1$. Then, from the divisibility of $1+3y=6x+4$ by $x$ and $x>1$, it follows that 4 is divisible by $x$ and the possibilities are $x=2,4$. Checking shows that the solution in this case is $x=4, y=9$.

Now consider the case $x \geq y>1$. From the divisibility of $1+3y \leq 1+3(x-1)=3x-2$ by $x$, it follows that $1+3y=x$ or $1+3y=2x$. If $1+3y=x$, then $1+2x=6y+3$ is divisible by $y$, so 3 is divisible by $y$ and $x=10, y=3$ is a solution to the problem.

If $1+3y=2x$, then $y$ is odd, $y=2k+1, k>0, x=3k+2$. Then $1+2x=6k+5$ must be divisible by $y=2k+1$, which means $6k+5-3(2k+1)=2$ is divisible by $y=2k+1 \geq 3$, which is impossible.

Evaluation Criteria. Only the correct answer: with verification: 1 point. Noted that $y$ cannot be divisible by $x$ and $x, y>1$: 1 point. Correctly found only one solution: 4 points.