AMC 10 · 2019 · #18

Grade 8 rate-ratio

Archetype Small-Domain Constrained Search

base-conversionsequences-geometriclinear-diophantinedigit-constraints convert-to-algebrasystematic-enumeration ↑ Prerequisites: base-conversionsequences-geometric

📏 Short solution 💡 2 insights

Problem

For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$ ?

$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: In some base $k$, the fraction $\frac{7}{51}$ written in base-$k$ "decimal" notation is the repeating expansion $0.232323\ldots_k$. Find the positive integer $k$.

Givens: $0.\overline{23}_k = 0.232323\ldots_k$ in base $k$; Its base-ten value is $\frac{7}{51}$; Choices: (A) $13$, (B) $14$, (C) $15$, (D) $16$, (E) $17$; Digits $2$ and $3$ must be valid in base $k$, so $k \geq 4$

Unknowns: The integer base $k$

Understand

Restated: In some base $k$, the fraction $\frac{7}{51}$ written in base-$k$ "decimal" notation is the repeating expansion $0.232323\ldots_k$. Find the positive integer $k$.

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #6 Guess and Check, #13 Convert to Algebra

Tool #3 (Eliminate): there are only $5$ choices for $k$, and each gives a number we can plug in and compare to $\frac{7}{51}$. Tool #6 (Guess and Check): test choices directly — the test is one arithmetic check per choice. Tool #13 (Algebra): set up the equation $\frac{2k+3}{k^2-1} = \frac{7}{51}$ to give us the right object to test.

Execute — Answer: D

#13 Convert to Algebra 8.EE.A.1 Step 1

Convert the repeating base-$k$ expansion to a fraction.
In base $k$, $0.\overline{23}_k = \frac{2k+3}{k^2} + \frac{2k+3}{k^4} + \cdots$ This is a geometric series with first term $\frac{2k+3}{k^2}$ and ratio $\frac{1}{k^2}$, so the sum is $\frac{(2k+3)/k^2}{1 - 1/k^2} = \frac{2k+3}{k^2-1}$.

$$0.\overline{23}_k = \dfrac{2k+3}{k^2-1}$$

💡 Same idea as $0.\overline{ab}_{10} = \frac{10a+b}{99}$ — replace $10$ by $k$ and $99$ by $k^2-1$.

#13 Convert to Algebra 8.EE.C.7 Step 2

Set this equal to $\frac{7}{51}$ and cross-multiply: $51(2k+3) = 7(k^2-1)$, which expands to $102k + 153 = 7k^2 - 7$.
Rearrange: $7k^2 - 102k - 160 = 0$.

$$7k^2 - 102k - 160 = 0$$

💡 Cross-multiply equals to clear fractions, then collect to a single quadratic.

#3 Eliminate Possibilities 6.EE.B.5 Step 3

Now use the multiple-choice gift: just plug each of $13, 14, 15, 16, 17$ into $7k^2 - 102k - 160$ and stop at the one that gives $0$.
Try $k = 16$: $7 \cdot 256 - 102 \cdot 16 - 160 = 1792 - 1632 - 160 = 0$.
Hit on the first try (or second, after trying $k = 13$ which gives $7 \cdot 169 - 102 \cdot 13 - 160 = 1183 - 1326 - 160 = -303 \ne 0$).

$$k = 16: \;\; 7(256) - 102(16) - 160 = 1792 - 1632 - 160 = 0 \checkmark$$

💡 Five choices, one equation, one plug-in each — straight elimination.

#6 Guess and Check 4.NF.A.1 Step 4

Double-check by computing $0.\overline{23}_{16}$ directly: $\frac{2 \cdot 16 + 3}{16^2 - 1} = \frac{35}{255} = \frac{7}{51}$ (divide top and bottom by $5$).
Match.
The answer is choice (D) $16$.

$$\dfrac{35}{255} = \dfrac{7}{51} \;\Rightarrow\; \textbf{(D) } 16$$

💡 Simplify $35/255$ by dividing by $\gcd = 5$ — lands exactly on $7/51$.

[1] #13 8.EE.A.1 Convert the repeating base-$k$ expansion to a fraction. In base $k$, $0.\overlin

[2] #13 8.EE.C.7 Set this equal to $\frac{7}{51}$ and cross-multiply: $51(2k+3) = 7(k^2-1)$, whic

[3] #3 6.EE.B.5 Now use the multiple-choice gift: just plug each of $13, 14, 15, 16, 17$ into $7

[4] #6 4.NF.A.1 Double-check by computing $0.\overline{23}_{16}$ directly: $\frac{2 \cdot 16 + 3

Review

Reasonableness: $\frac{7}{51} \approx 0.1373$. In base $16$, $0.\overline{23}_{16} = \frac{35}{255} \approx 0.1373$ — matches. We could also note that $51 = 3 \cdot 17$ and $\gcd(2k+3, k^2-1)$ needs to leave denominator $51$ after reduction; $k = 16$ gives $\frac{35}{255}$ which simplifies via $\gcd = 5$ to $\frac{7}{51}$. Other choices: $k = 13 \to \frac{29}{168}$ (no), $k = 14 \to \frac{31}{195}$ (no), $k = 15 \to \frac{33}{224}$ (no), $k = 17 \to \frac{37}{288}$ (no). Only (D) works.

Alternative: Tool #15 (Reorganize): factor $7k^2 - 102k - 160 = (7k + ?)(k - ?)$. Trying $(7k + 10)(k - 16) = 7k^2 - 112k + 10k - 160 = 7k^2 - 102k - 160$ — matches. So $k = 16$ (positive root) or $k = -10/7$ (rejected). Same answer, no choice-checking needed.

CCSS standards used (min grade 8)

4.NF.A.1 Explain why a fraction is equivalent to another fraction (Simplifying $\frac{35}{255}$ to $\frac{7}{51}$ by dividing by $\gcd = 5$.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Testing each candidate $k \in \{13, \ldots, 17\}$ against the equation.)
8.EE.A.1 Know and apply the properties of integer exponents (Summing the geometric series $\sum k^{-2n}$ to get the closed-form $\frac{2k+3}{k^2-1}$.)
8.EE.C.7 Solve linear equations in one variable (Rearranging $51(2k+3) = 7(k^2-1)$ to the standard quadratic form.)

⭐ This AMC 10 problem only needs Grade 8 equation-solving you already know — convert the repeating base-$k$ digits to $\frac{2k+3}{k^2-1}$, set it equal to $\frac{7}{51}$, and plug each choice in. Only $k = 16$ fits. The answer is $\textbf{(D)}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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