2022 AMC 10A #17

Problem:

How many three-digit positive integers $\underset{―}{a}$ $\underset{―}{b}$ $\underset{―}{c}$ are there whose nonzero digits $a$, $b$, and $c$ satisfy

$$0.\overline{\underset{―}{a}\underset{―}{b}\underset{―}{c}}=\frac{1}{3}(0.\bar{a}+0.\bar{b}+0.\bar{c})?$$(The bar indicates repetition, thus $0.\overline{\underset{―}{a}\underset{―}{b}\underset{―}{c}}$ in the infinite repeating decimal $0.\underset{―}{a}\underset{―}{b}\underset{―}{c}\underset{―}{a}\underset{―}{b}\underset{―}{c}\cdots$)

$\mathbf{\text{(A) }}9\mathbf{\text{(B) }}10\mathbf{\text{(C) }}11\mathbf{\text{(D) }}13\mathbf{\text{(E) }}14$

Solution:

Note that we can rewrite the condition as $$\frac{\overline{abc}}{999}=\frac{1}{3}(\frac{a}{9}+\frac{b}{9}+\frac{c}{9})$$

$$⟹\overline{abc}=37(a+b+c)$$

$$⟹100a+10b+c=37a+37b+37c$$

$$⟹63a-27b-36c=0$$

$$⟹7a=3b+4c$$

Now, after casework on $a=1,2,3,\dots ,9$, we get that there are $\boxed{{\displaystyle \mathbf{\text{13 (D)}}}}$ such three-digit integers $\overline{abc}$.