2. Find the smallest natural number $m$ such that $28 m+13=p d$ and $m-71=q d$, where $p, q$, and $d$ are natural numbers, $q \neq 1$ and $d>3$.
III/2. From the equation $m-71=q d$ we express $m$ and substitute it into the equation $28 m+13=p d$. We get $28 q d+2001=p d$, so $d \mid 2001$. Since $2001=1 \cdot 3 \cdot 23 \cdot 29$ and $d>3$, we have $d=23$. For $q$ we must take 2 to get the smallest $m$. Thus we get $q d=2 \cdot 23=m-71$ and $m=117$.

Eliminating $m$ from the system and obtaining the equation $28 q d+2001=p d$: $\mathbf{2}$ points. Conclusion $d \mid 2001$: 1 point. Conclusion $d=23$: 1 point. Conclusion “the smallest $m$ will be for $q=2$”: 2 points. Conclusion $m=117$: $\mathbf{1}$ point. If the contestant guesses the number $m$: 2 points.