2022 AMC 10A Problem 17

The repeating decimals satisfy $$\begin{aligned} 0.\overline{\underline{a}\underline{b}\underline{c}}&=\frac{100a+10b+c}{999},\\ 0.\overline a&=\frac a9,\quad 0.\overline b=\frac b9,\\ 0.\overline c&=\frac c9. \end{aligned}$$ Substitution and multiplication by $999$ give $$100a+10b+c=37(a+b+c),$$ or $$7a=3b+4c.$$ For each $a\in\{1,2,\ldots,9\},$ checking the nonzero digits $b,c$ in this linear equation gives $$\begin{gathered} (1,1,1),(2,2,2),(3,3,3),\\ (4,4,4),(4,8,1),(5,1,8),\\ (5,5,5),(5,9,2),(6,2,9),\\ (6,6,6),(7,7,7),(8,8,8),\\ (9,9,9). \end{gathered}$$ Thus there are $13$ integers. Thus, D is the correct solution.