AMC 10 · 2019 · #11

Grade 8 number-theory
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, the fraction seven fifty-firsts written in that base's decimal notation is the repeating expansion zero point two three two three two three, forever. Find that positive integer base.

Pick an answer.

(A)
13
(B)
14
(C)
15
(D)
16
(E)
17
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 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 (2k+3)/(k²-1) = 7/51 to give us the right object to test.

1STEP 1

Repeating expansion as a fraction

A two-digit cycle makes the denominator the square minus one.

0.23_k = (2k+3)/(k²-1)
2STEP 2

Set up the quadratic

Cross-multiplying gives a quadratic.

7k² - 102k - 160 = 0
3STEP 3

Find the integer root

Testing integers finds the one that works.

k = 16: 7(256) - 102(16) - 160 = 1792 - 1632 - 160 = 0 ✓
4STEP 4

Check it back

Reducing returns the original fraction, so the base is 16.

35/255 = 7/51 → (D) 16
Answer
16
7/51 ≈ 0.1373. In base 16, 0.23₁₆ = 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 35/255 which simplifies via gcd = 5 to 7/51. Other choices: k = 13 → 29/168 (no), k = 14 → 31/195 (no), k = 15 → 33/224 (no), k = 17 → 37/288 (no). Only (D) works.
💡Key takeaway

This AMC 12 problem only needs Grade 8 equation-solving you already know — convert the repeating base-k digits to (2k+3)/(k²-1), set it equal to 7/51, and plug each choice in. Only k = 16 fits. The answer is (D).

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