5.1. Find the smallest natural number $n$ such that the natural number $n^{2}+14 n+13$ is divisible by 68.
Answer: 21

Solution. Note that $n^{2}+14 n+13=(n+1)(n+13)$, and $68=4 \cdot 17$. Therefore, one of the numbers $n+1$ and $n+13$ must be divisible by 17. Moreover, $n$ must be odd; otherwise, $n+1$ and $n+13$ would both be odd. The smallest such natural number $n$ is $2 \cdot 17-13=21$. Note that it works because $21^{2}+14 \cdot 21+13=$ $(21+1) \cdot(21+13)=22 \cdot 34=11 \cdot 68$.