2022 AMC 10A Problem 17

Convert the repeating decimals: $0.\overline{abc} = \dfrac{100a + 10b + c}{999}$ and $0.\overline{a} = \dfrac{a}{9}$, so the right side is $\dfrac{a + b + c}{27}$. The condition becomes
$$\frac{100a + 10b + c}{999} = \frac{a+b+c}{27} \quad\Longrightarrow\quad 100a + 10b + c = 37(a + b + c),$$
which simplifies to $63a = 27b + 36c$, or $7a = 3b + 4c$.

Rewrite as $3(a - b) = 4(c - a)$. Since $\gcd(3,4) = 1$, $a - b$ is a multiple of $4$: $a - b = 4t$ and $c - a = 3t$ for some integer $t$, so
$$b = a - 4t, \qquad c = a + 3t.$$
Now count digit solutions with $1 \le a, b, c \le 9$:

• $t = 0$: $a = b = c$, any of $9$ values. $9$ numbers ($111, 222, \dots, 999$).
• $t = 1$: $b = a - 4 \ge 1$ needs $a \ge 5$; $c = a + 3 \le 9$ needs $a \le 6$. So $a = 5, 6$: $518$ and $629$. $2$ numbers.
• $t = -1$: $b = a + 4 \le 9$ needs $a \le 5$; $c = a - 3 \ge 1$ needs $a \ge 4$. So $a = 4, 5$: $481$ and $592$. $2$ numbers.
• $|t| \ge 2$: the conditions $a \ge 4|t| + 1$ and $a \le 9 - 3|t|$ (or the mirror pair) are incompatible.

Total: $9 + 2 + 2 = 13$.

The answer is $\boxed{\textbf{(D)}\ 13}$.

Repeating decimals are fractions with denominators $9, 99, 999$, so the equation is really a linear Diophantine equation in the digits. Parametrizing all integer solutions of $3(a-b) = 4(c-a)$ with a single integer $t$ turns a three-variable search into a short check of a few $t$ values against the digit bounds.

Casework on $a$ in $7a = 3b + 4c$: for each $a$, $c$ is determined mod $3$ by $4c \equiv 7a \pmod 3$, and only a few $c$ values keep $b$ between $1$ and $9$. The counts by $a = 1, \dots, 9$ are $1, 1, 1, 2, 3, 2, 1, 1, 1$, again summing to $13$.

• Counting only the nine solutions with a = b = c, or allowing b or c to be 0 (all three digits must be nonzero).
• Miscounting the $t = \pm 1$ cases by checking only one of the two digit bounds (for example allowing $a = 7$ with $t = 1$, where $c = 10$).