7.8. Determine how many two-digit numbers $\overline{a b}$ satisfy the following conditions simultaneously:
a) $\overline{a b}$ is not divisible by 37;

b) the number $\overline{a b a b a b \ldots a b}$, where $\overline{a b}$ appears 2019 times, is divisible by 37.
## Solution:

From the fact that $\overline{a b}$ is not divisible by $37 \Rightarrow \overline{a b} \notin\{37 ; 74\}$. Consider the number $n=\overline{a b a b a b \ldots a b}$, where $\overline{a b}$ appears 2019 times. Then, $n=\overline{a b} \cdot\left(10^{4036}+10^{4034}+10^{4032}+\cdots+10^{2}+1\right)$. Since $10^{4036}+10^{4034}+10^{4032}+\cdots+10^{4}+10^{2}+1=\left(1+10^{2}+10^{4}\right)+\left(10^{6}+10^{8}+10^{10}\right)+\cdots+\left(10^{4032}+10^{4034}+10^{4036}\right)=10101+10101 \cdot 10^{6}+10101 \cdot 10^{12}+\cdots+10101 \cdot 10^{4032}=10101 \cdot\left(1+10^{6}+10^{12}+\cdots+10^{4032}\right)$, we obtain that $n=\overline{a b} \cdot 10101 \cdot\left(1+10^{6}+10^{12}+\cdots+10^{4032}\right)$. But $10101 \vdots 37 \Rightarrow n \vdots 37$, regardless of $\overline{a b}$. Thus, the numbers $\overline{a b} \in \mathbb{N} \backslash\{37 ; 74\}$. Therefore, there are 88 such numbers.