AMC 10 · 2019 · #18

Grade 8 rate-ratio

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

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

📏 Short solution 💡 2 insights

Problem

In some base k, the fraction

\frac{7}{51}

written in base-k "decimal" notation is the repeating expansion 0.232323…_k. Find the positive integer k.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

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 $\frac{2k+3}{k²-1}$ = $\frac{7}{51}$ to give us the right object to test.

1STEP 1

The repeating base-k digits form a geometric series that sums to $\frac{2k+3}{k²-1}$ .

0.23_k =

\frac{2k+3}{k²-1}

2STEP 2

Setting it equal to $\frac{7}{51}$ and cross-multiplying clears the fractions into the quadratic 7k² - 102k - 160 = 0.

7k² - 102k - 160 = 0

3STEP 3

Plug each choice into the quadratic; only k = 16 makes it zero (1792 - 1632 - 160 = 0).

k = 16: 7(256) - 102(16) - 160 = 1792 - 1632 - 160 = 0 ✓

4STEP 4

Check directly: 0.23 in base 16 is $\frac{35}{255}$ , which reduces by 5 to $\frac{7}{51}$ — confirming the answer.

\frac{35}{255}

\frac{7}{51}

→ (D) 16

Answer

\frac{7}{51}

≈ 0.1373. In base 16, 0.23₁6 =

\frac{35}{255}

≈ 0.1373 — matches. We could also note that 51 = 3 · 17 and gcd(2k+3, k²-1) 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 →

\frac{29}{168}

(no), k = 14 →

\frac{31}{195}

(no), k = 15 →

\frac{33}{224}

(no), k = 17 →

\frac{37}{288}

(no). Only (D) works.

💡Key takeaway

This AMC 10 problem only needs Grade 8 equation-solving you already know — convert the repeating base-k digits to $\frac{2k+3}{k²-1}$ , set it equal to $\frac{7}{51}$ , and plug each choice in. Only k = 16 fits. The answer is (D).

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