AMC 10 · 2021 · #8

Grade 8 arithmetic

Archetype Small-Domain Constrained Search

fraction-decimal-conversiondigit-decompositionlinear-equations-one-var convert-to-algebraguess-and-check ↑ Prerequisites: fraction-decimal-conversion

📏 Medium solution 💡 2 insights

Problem

When a student multiplied the number $66$ by the repeating decimal,
$\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},$
where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$

$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: A student should have multiplied $66$ by the repeating decimal $1.\overline{ab}$ where $a$ and $b$ are digits. Instead the student used the terminating decimal $1.ab$ (just two digits after the point). The wrong answer turned out to be $0.5$ less than the correct one. Find the two-digit number $\overline{ab}$.

Givens: Correct multiplier: $1.\overline{ab} = 1.ababab\ldots$; Student's multiplier: $1.ab$ (only two decimal digits); Both multipliers are multiplied by $66$; Correct $-$ wrong $= 0.5$; Choices: (A) $15$, (B) $30$, (C) $45$, (D) $60$, (E) $75$

Unknowns: The two-digit number $\overline{ab}$ (a value from $10$ to $99$, possibly $00$-$09$ as well)

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #3 Eliminate Possibilities, #11 Work Backwards

Only five candidate values $\{15, 30, 45, 60, 75\}$ — Tool #6 (Guess and Check) plugs each into the repeating-vs-terminating gap and tests whether $66 \times (\text{gap}) = 0.5$. Tool #3 (Eliminate) discards every $N$ that misses $0.5$. Tool #11 (Work Backwards) is the verification: starting from the required gap $\dfrac{0.5}{66} = \dfrac{1}{132}$, run the gap formula in reverse to recover $N$ in one step — this confirms the unique answer without sweeping all five choices.

Execute — Answer: E

#11 Work Backwards 8.NS.A.1 Step 1

Write both decimals as fractions of the two-digit number $N = 10a + b$.
The terminating $1.ab = 1 + \dfrac{N}{100}$ (two decimal places means denominator $100$).
The repeating $1.\overline{ab} = 1 + \dfrac{N}{99}$ (the standard rule: a $2$-digit block repeats forever $\Rightarrow$ denominator $10^2 - 1 = 99$).

$$1.ab = 1 + \dfrac{N}{100},\qquad 1.\overline{ab} = 1 + \dfrac{N}{99}$$

💡 Two-place terminating $\Rightarrow$ /$100$; two-place repeating $\Rightarrow$ /$99$. Standard Grade 8 conversion.

#11 Work Backwards 5.NF.A.1 Step 2

Compute the gap between correct and wrong multipliers.
The $1$s cancel.
Find a common denominator $9900$.

$$1.\overline{ab} - 1.ab = \dfrac{N}{99} - \dfrac{N}{100} = \dfrac{100N - 99N}{9900} = \dfrac{N}{9900}$$

💡 The two fractions differ by exactly $\dfrac{N}{9900}$ — a small number because the missed repeat is tiny.

#11 Work Backwards 6.NS.B.3 Step 3

Multiply the gap by $66$ to get the gap between the correct and wrong products.
Simplify $\dfrac{66}{9900} = \dfrac{1}{150}$.

$$66 \cdot \dfrac{N}{9900} = \dfrac{66 N}{9900} = \dfrac{N}{150}$$

💡 $9900 = 150 \times 66$, so $66/9900$ collapses to $1/150$.

#11 Work Backwards 6.EE.B.7 Step 4

The problem says this gap equals $0.5$.
Solve for $N$.

$$\dfrac{N}{150} = 0.5\;\Rightarrow\;N = 150 \cdot 0.5 = 75$$

💡 Running the formula backwards: multiply both sides by $150$.

#6 Guess and Check 7.NS.A.3 Step 5

Verify with Tool #6 — plug $N = 75$ back.
Correct: $66 \cdot 1.\overline{75} = 66 \cdot \dfrac{174}{99} = \dfrac{11484}{99} = 116.0\overline{6}\ldots$, namely $\dfrac{11484}{99}$.
Wrong: $66 \cdot 1.75 = 115.5$.
Difference: $\dfrac{11484}{99} - 115.5 = \dfrac{11484 - 11434.5}{99} = \dfrac{49.5}{99} = 0.5$ ✓.

$$66 \cdot 1.\overline{75} - 66 \cdot 1.75 = 0.5\;\checkmark$$

💡 Plug the candidate back into the original word relation — must hit $0.5$ exactly.

#3 Eliminate Possibilities 4.NBT.A.2 Step 6

$N = 75$ matches choice (E).

$$N = 75\;\Rightarrow\;\textbf{(E)}$$

💡 Read the two-digit value off the five options.

[1] #11 8.NS.A.1 Write both decimals as fractions of the two-digit number $N = 10a + b$. The term

[2] #11 5.NF.A.1 Compute the gap between correct and wrong multipliers. The $1$s cancel. Find a c

[3] #11 6.NS.B.3 Multiply the gap by $66$ to get the gap between the correct and wrong products.

[4] #11 6.EE.B.7 The problem says this gap equals $0.5$. Solve for $N$.

[5] #6 7.NS.A.3 Verify with Tool #6 — plug $N = 75$ back. Correct: $66 \cdot 1.\overline{75} = 6

[6] #3 4.NBT.A.2 $N = 75$ matches choice (E).

Review

Reasonableness: Each given choice corresponds to a different gap $\dfrac{N}{150}$ which would have to equal $0.5$. Only $N = 75$ works — the other four choices give gaps $\dfrac{15}{150} = 0.1,\;\dfrac{30}{150} = 0.2,\;\dfrac{45}{150} = 0.3,\;\dfrac{60}{150} = 0.4$ — none of which is $0.5$. The unique value is (E). The size also makes sense: a missed two-digit repeat is small, so the resulting product gap of $0.5$ on a multiplier near $1$ is plausible.

Alternative: Tool #6 (pure Guess and Check) by typing each choice directly. $N=15$: gap $= 0.1$, miss. $N=30$: $0.2$, miss. $N=45$: $0.3$, miss. $N=60$: $0.4$, miss. $N=75$: $0.5$, hit — (E). Less elegant than the backwards path but completely mechanical and requires no algebra.

CCSS standards used (min grade 8)

8.NS.A.1 Know that numbers that are not rational are called irrational numbers (Converting the repeating decimal $1.\overline{ab}$ into the rational form $1 + \frac{N}{99}$ — the standard repeating-decimal-to-fraction rule taught in Grade 8.)
5.NF.A.1 Add and subtract fractions with unlike denominators (Computing the gap $\frac{N}{99} - \frac{N}{100}$ via the common denominator $9900$.)
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals (Simplifying $\frac{66}{9900}$ to $\frac{1}{150}$ and arithmetic with $0.5$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving $\frac{N}{150} = 0.5$ for $N$.)
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers (Verifying that $66 \cdot 1.\overline{75} - 66 \cdot 1.75 = 0.5$.)
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Matching the computed value $75$ to choice (E).)

⭐ This AMC 10 problem only needs Grade 8 repeating-decimal-to-fraction conversion you already know — turn $1.\overline{ab} = 1 + \frac{N}{99}$ and $1.ab = 1 + \frac{N}{100}$, find their gap $\frac{N}{9900}$, multiply by $66$ to get $\frac{N}{150} = 0.5$, and $N = 75$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 8)

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