AMC 10 · 2019 · #18
Grade 8 rate-ratioProblem
For some positive integer k, the repeating base-k representation of the (base-ten) fraction 517 is 0.23k=0.232323...k. What is k?
Pick an answer.
AMC 10 2019 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.
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 $k$.
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 $k \geq 4$
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
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 $\frac{(2k+3)/k^2}{1 - 1/k^2} = \frac{2k+3}{k^2-1}$.
💡 Same idea as $0.\overline{ab}_{10} = \frac{10a+b}{99}$ — replace $10$ by $k$ and $99$ by $k^2-1$.
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: $7k^2 - 102k - 160 = 0$.
💡 Cross-multiply equals to clear fractions, then collect to a single quadratic.
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 $7 \cdot 169 - 102 \cdot 13 - 160 = 1183 - 1326 - 160 = -303 \ne 0$).
💡 Five choices, one equation, one plug-in each — straight elimination.
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) $16$.
💡 Simplify $35/255$ by dividing by $\gcd = 5$ — lands exactly on $7/51$.
8.EE.A.1 Convert the repeating base-$k$ expansion to a fraction. In base $k$, $0.\overlin 8.EE.C.7 Set this equal to $\frac{7}{51}$ and cross-multiply: $51(2k+3) = 7(k^2-1)$, whic 6.EE.B.5 Now use the multiple-choice gift: just plug each of $13, 14, 15, 16, 17$ into $7 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.1Explain why a fraction is equivalent to another fraction (Simplifying $\frac{35}{255}$ to $\frac{7}{51}$ by dividing by $\gcd = 5$.)6.EE.B.5Understand 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.1Know 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.7Solve 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 $\textbf{(D)}$.
⭐ 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 $\textbf{(D)}$.
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