AMC 10 · 2022 · #17

Grade 7 arithmetic

fraction-decimal-conversiondigit-decompositiondivisibility-rules convert-to-algebrasystematic-enumeration ↑ Prerequisites: fraction-decimal-conversion

📏 Medium solution 💡 3 insights

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 $0.\overline{\underline{a}~\underline{b}~\underline{c}}$ is the infinite repeating decimal $0.\underline{a}~\underline{b}~\underline{c}~\underline{a}~\underline{b}~\underline{c}~\cdots$ )

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Count three-digit positive integers $\overline{abc}$ whose digits $a, b, c$ are all nonzero and satisfy $0.\overline{abc} = \tfrac{1}{3}(0.\overline{a} + 0.\overline{b} + 0.\overline{c})$, where the bars mean infinite repetition.

Givens: $a, b, c \in \{1, 2, \dots, 9\}$ (each digit nonzero); $0.\overline{a} = \tfrac{a}{9}$, $0.\overline{b} = \tfrac{b}{9}$, $0.\overline{c} = \tfrac{c}{9}$; $0.\overline{abc} = \tfrac{100a + 10b + c}{999}$; Choices: (A) $9$, (B) $10$, (C) $11$, (D) $13$, (E) $14$

Unknowns: Number of ordered triples $(a, b, c)$ satisfying the equation

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #2 Make a Systematic List, #9 Solve an Easier Related Problem, #3 Eliminate Possibilities

First reduce the decimal equation to an integer relation. Repeating-decimal-to-fraction conversion (Tool #13 Convert to Algebra) turns the unwieldy equation into $7a = 3b + 4c$. Once we have a clean equation in three nonzero digits, Tool #2 (Systematic List) sweeps every value of $a$ from $1$ to $9$ and counts $(b, c)$ pairs satisfying $3b + 4c = 7a$ with $b, c \in \{1, \dots, 9\}$. Tool #9 (Easier Problem) shows up as the observation: trying $a=b=c$ first reveals a whole family of solutions and orients the search.

Execute — Answer: D

#13 Convert to Algebra 7.NS.A.2 Step 1

Convert each repeating decimal to a fraction.
A single repeating digit $d$ gives $0.\overline{d} = \tfrac{d}{9}$.
A three-digit repeating block $\overline{abc}$ gives $0.\overline{abc} = \tfrac{100a+10b+c}{999}$.
Substituting into the equation and using $999 = 27 \cdot 37$:

$$\dfrac{100a+10b+c}{999} = \dfrac{1}{3} \cdot \dfrac{a+b+c}{9} = \dfrac{a+b+c}{27}$$

💡 Every repeating decimal is a fraction with $9, 99, 999, \dots$ in the denominator — that's the only fact about repeating decimals we need.

#13 Convert to Algebra 7.EE.A.1 Step 2

Multiply both sides by $999 = 37 \cdot 27$.
The right side becomes $37(a+b+c)$.
Expand and group like terms:

$$100a + 10b + c = 37(a+b+c) \;\Longrightarrow\; 63a = 27b + 36c \;\Longrightarrow\; 7a = 3b + 4c$$

💡 Divide through by the common factor $9$ — what was a fancy decimal equation becomes a single tidy linear relation in $a, b, c$.

#9 Solve an Easier Related Problem 6.EE.B.6 Step 3

Sub-case I: try $a = b = c$.
Then $7a = 3a + 4a = 7a$ holds for every digit $a$.
So all nine triples $(1,1,1), (2,2,2), \dots, (9,9,9)$ are solutions — that's $9$ triples right away.

$(a,a,a)$ for $a \in \{1, \dots, 9\}$ — $9$ solutions.

💡 Try the simplest symmetric guess first — sometimes the equation collapses to an identity.

#2 Make a Systematic List 6.NS.B.4 Step 4

Sub-case II: list non-diagonal solutions systematically.
For each $a$ from $1$ to $9$ scan $b \in \{1, \dots, 9\}$ and check whether $c = \tfrac{7a - 3b}{4}$ is an integer in $\{1, \dots, 9\}$ different from $a$ (or with $(b,c) \neq (a,a)$).
The key observation: $4c = 7a - 3b$ forces $7a - 3b \equiv 0 \pmod{4}$, i.e.
$3a + b \equiv 0 \pmod 4$ (since $7 \equiv 3, -3 \equiv 1 \pmod 4$).

$c = \dfrac{7a - 3b}{4}$, scan $a = 1, \dots, 9$ and $b = 1, \dots, 9$.

💡 Once $a$ and $b$ are fixed, $c$ is forced; checking divisibility and range is mechanical.

#2 Make a Systematic List 6.EE.B.5 Step 5

Carry out the scan.
The off-diagonal solutions found are $(a,b,c) = (4,8,1), (5,1,8), (5,9,2), (6,2,9)$.
Quick verification: $7 \cdot 4 = 28 = 3 \cdot 8 + 4 \cdot 1$; $7 \cdot 5 = 35 = 3 \cdot 1 + 4 \cdot 8 = 3 + 32$; $7 \cdot 5 = 35 = 3 \cdot 9 + 4 \cdot 2 = 27 + 8$; $7 \cdot 6 = 42 = 3 \cdot 2 + 4 \cdot 9 = 6 + 36$.
All four check out.

Off-diagonal: $\{(4,8,1), (5,1,8), (5,9,2), (6,2,9)\}$ — $4$ solutions.

💡 Verify each candidate by plugging back into $7a = 3b + 4c$.

#3 Eliminate Possibilities 4.OA.A.3 Step 6

Total count: $9 + 4 = 13$ ordered triples, each giving a distinct three-digit integer $\overline{abc}$.
Match to the choices.

$$9 + 4 = 13 \;\Rightarrow\; \textbf{(D)}$$

💡 Add the diagonal and off-diagonal counts; pick the matching choice.

[1] #13 7.NS.A.2 Convert each repeating decimal to a fraction. A single repeating digit $d$ gives

[2] #13 7.EE.A.1 Multiply both sides by $999 = 37 \cdot 27$. The right side becomes $37(a+b+c)$.

[3] #9 6.EE.B.6 Sub-case I: try $a = b = c$. Then $7a = 3a + 4a = 7a$ holds for every digit $a$.

[4] #2 6.NS.B.4 Sub-case II: list non-diagonal solutions systematically. For each $a$ from $1$ t

[5] #2 6.EE.B.5 Carry out the scan. The off-diagonal solutions found are $(a,b,c) = (4,8,1), (5,

[6] #3 4.OA.A.3 Total count: $9 + 4 = 13$ ordered triples, each giving a distinct three-digit in

Review

Reasonableness: Did the off-diagonal search miss anything? Bounds: $b, c \in \{1, \dots, 9\}$ gives $3b + 4c \in \{7, \dots, 63\}$, so $7a \in \{7, \dots, 63\}$ allows $a \in \{1, \dots, 9\}$ — all in range. The shift identity $(a, b, c) \to (a, b+4, c-3)$ preserves $3b + 4c$, so from each diagonal $(a,a,a)$ we can step at most a few times before leaving $\{1, \dots, 9\}$. Stepping up from $(4,4,4)$ reaches $(4,8,1)$ (valid) then $(4,12,-2)$ (invalid). Stepping up from $(5,5,5)$ reaches $(5,9,2)$ (valid); stepping down reaches $(5,1,8)$ (valid). Stepping down from $(6,6,6)$ reaches $(6,2,9)$ (valid). Stepping from $(1,1,1), (2,2,2), (3,3,3), (7,7,7), (8,8,8), (9,9,9)$ leaves the digit range immediately. Total off-diagonal: exactly $4$, confirming the count.

Alternative: Tool #5 (Look for a Pattern) on the shift map $(b, c) \to (b+4, c-3)$: since $\gcd(3, 4) = 1$, every solution for fixed $a$ lies on a single arithmetic chain $(b, c), (b \pm 4, c \mp 3), \dots$ within the $9 \times 9$ grid. Counting the chain length per $a$ gives the same total $13$ without per-cell case-checks.

CCSS standards used (min grade 7)

7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers (Converting the repeating decimals $0.\overline{d}$ and $0.\overline{abc}$ into the fractions $\tfrac{d}{9}$ and $\tfrac{100a+10b+c}{999}$.)
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions (Multiplying through by $999$, expanding $37(a+b+c)$, and simplifying $63a = 27b + 36c$ to $7a = 3b + 4c$.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Testing the symmetric guess $a = b = c$ to identify the diagonal family of solutions.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Using divisibility by $4$ to decide when $\tfrac{7a - 3b}{4}$ is an integer during the scan.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Verifying that each candidate triple $(a,b,c)$ actually satisfies $7a = 3b + 4c$.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Adding the diagonal count ($9$) and off-diagonal count ($4$) to get the final total.)

⭐ Every repeating decimal is just a fraction over $9$, $99$, $999, \dots$. Convert and simplify, and the messy decimal equation collapses to $7a = 3b + 4c$. The nine diagonal triples $(a, a, a)$ always work; a quick scan over $a$ turns up four more — $(4,8,1), (5,1,8), (5,9,2), (6,2,9)$ — for a total of $\textbf{(D) }13$.