74th Romanian Mathematical Olympiad

a) If $d$ is a common divisor of the numbers $12n+13$ and $13n+14$, then $d$ divides the numbers $13(12n+13)$ and $12(13n+14)$. Then $d$ divides $13(12n+13)-12(13n+14)$, that is $d\mid 1$. So $d=1$, hence the numbers $12n+13$ and $13n+14$ are coprime.

b) The relation $\frac{a}{b}=\frac{12n+13}{13n+14}$ leads to $a(13n+14)=b(12n+13)$, hence $13n+14\mid b(12n+13)$. Since $12n+13$ and $13n+14$ are coprime, $13n+14$ divides $b$. Since

$b\ne 0$ (it is the denominator of a fraction), there exists $k\in {\mathbb{N}}^{\ast }$ so that $b=k(13n+14)$ and therefore $a=k(12n+13)$.

Replace $a=k(12n+13)$ and $b=k(13n+14)$ in $17a+19b<2024$ to get $k(451n+487)<2024$. Then $451n+487<2024$, whence $n\le 3$.

For $n=0$ we get $k\le 4$, so $(a,b)\in \{(13,14),(26,28),(39,42),(52,56)\}$. For $n=1$ we get $k\le 2$, so $(a,b)\in \{(25,27),(50,54)\}$. For $n\in \{2,3\}$ we get $k=1$, so $(a,b)\in \{(37,40),(49,53)\}$.

In total, there are 8 pairs.