2022 AMC 10A Problem 17

Step 1: turn the repeating decimals into fractions. If $x = 0.\overline{abc}$, then $1000x = abc.\overline{abc}$, and subtracting gives $999x = \underline{a}\,\underline{b}\,\underline{c} = 100a + 10b + c$. So
$$0.\overline{abc} = \frac{100a + 10b + c}{999}.$$
In the same way $0.\overline{a} = \frac{a}{9}$ (for example $0.\overline{3} = \frac39 = \frac13$). So the right side is
$$\frac13 \cdot \frac{a + b + c}{9} = \frac{a + b + c}{27}.$$

Step 2: simplify the equation. We need $\dfrac{100a + 10b + c}{999} = \dfrac{a + b + c}{27}$. Since $999 = 27 \times 37$, multiply both sides by $999$:
$$100a + 10b + c = 37a + 37b + 37c, \quad\text{so}\quad 63a = 27b + 36c.$$
Divide by $9$: $\ 7a = 3b + 4c$.

Step 3: rearrange to see the structure. Write $7a$ as $3a + 4a$:
$$3a + 4a = 3b + 4c \quad\Longrightarrow\quad 3(a - b) = 4(c - a).$$
The left side is a multiple of $3$, so $4(c - a)$ is too, which forces $c - a$ to be a multiple of $3$. Likewise $a - b$ must be a multiple of $4$. Digits are between $1$ and $9$, so $a - b$ is between $-8$ and $8$.

Step 4: try each possible value of $a - b$. Then $c - a = \frac34(a - b)$.

Check one: for $518$, $7 \cdot 5 = 35$ and $3 \cdot 1 + 4 \cdot 8 = 35$.

Step 5: add. $9 + 2 + 2 = 13$.

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

Repeating decimals are fractions over $9$, $99$, $999$, and so on, so the decimal condition is really an equation in whole-number digits. Rearranging it as $3(a - b) = 4(c - a)$ shows that the differences must be multiples of $4$ and $3$, which cuts a three-digit search down to a handful of cases.

Go through $a = 1, 2, \ldots, 9$ in $7a = 3b + 4c$ and look for digit pairs $b, c$ from $1$ to $9$. The counts for $a = 1$ to $9$ come out as $1, 1, 1, 2, 3, 2, 1, 1, 1$, again $13$ in total. For example, $a = 5$ needs $3b + 4c = 35$, which has the solutions $(b, c) = (1, 8), (5, 5), (9, 2)$.

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 $\boxed{\textbf{(D)}\ 13}$.