Problem:
How many three-digit positive integers $\underline{a}$ $\underline{b}$ $\underline{c}$ are there whose nonzero digits $a$, $b$, and $c$ satisfy
$$0.\overline{\underline{a}~\underline{b}~\underline{c}} = \frac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?$$(The bar indicates repetition, thus $0.\overline{\underline{a}~\underline{b}~\underline{c}}$ in the infinite repeating decimal $0.\underline{a}~\underline{b}~\underline{c}~\underline{a}~\underline{b}~\underline{c}~\cdots$)
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }13\qquad\textbf{(E) }14$
Solution:
Note that we can rewrite the condition as $$\frac{\overline{abc}}{999} = \frac{1}{3}\left(\frac{a}{9} + \frac{b}{9} + \frac{c}{9}\right)$$
$$\implies \overline{abc}=37(a+b+c)$$
$$\implies 100a+10b+c=37a+37b+37c$$
$$\implies 63a-27b-36c=0$$
$$\implies 7a=3b+4c$$
Now, after casework on $a=1, 2, 3, \dots, 9$, we get that there are $\boxed{\textbf{13 (D)}}$ such three-digit integers $\overline{abc}$.
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