B1. Find all pairs of natural numbers $(m, n)$ for which

$$
m+3 n-5=2 v-11 d
$$

where $v$ is the least common multiple of the numbers $m$ and $n$, and $d$ is the greatest common divisor of the numbers $m$ and $n$.

(6 points)
B1. Let's write $m=d a$ and $n=d b$. Thus, $a$ and $b$ are coprime natural numbers and $v=d a b$. If we substitute this into the equation, we get $d a + 3 d b - 5 = 2 d a b - 11 d$, so $d$ must divide 5. First, let's assume $d=1$. Then we can rearrange the equation to $a(2 b - 1) = 3 b + 6$. From this, it follows that $2 b - 1$ divides $3 b + 6$, and thus also divides $2(3 b + 6) - 3(2 b - 1) = 15$. Since $2 b - 1$ is a natural number, we have four possibilities. If $2 b - 1 = 1$, then $b = 1$ and $a = 9$. If $2 b - 1 = 3$, then $b = 2$ and $a = 4$, which is a contradiction since $a$ and $b$ are not coprime. If $2 b - 1 = 5$, then $b = 3$ and $a = 3$, which is again a contradiction. If $2 b - 1 = 15$, then $b = 8$ and $a = 2$, which is again a contradiction. In this case, we have the solution $(9,1)$. Now let $d=5$. Then we can divide the equation by 5 and rearrange it to $a(2 b - 1) = 3 b + 10$. From this, it follows that $2 b - 1$ divides $3 b + 10$, and thus also divides $2(3 b + 10) - 3(2 b - 1) = 23$. We have two possibilities. If $2 b - 1 = 1$, then $b = 1$ and $a = 13$. If $2 b - 1 = 23$, then $b = 12$ and $a = 2$, which is a contradiction since $a$ and $b$ are not coprime. Thus, we get another solution $(65,5)$.[^0]