2013 AMC 12A Problem 10

Attempt Problem 10 of the 2013 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

Let SS be the set of positive integers nn for which 1n\dfrac{1}{n} has the repeating decimal representation 0.ab‾=0.ababab…,0.\overline{ab} = 0.ababab\ldots, with aa and bb different digits. What is the sum of the elements of S?S?
1111
4444
110110
143143
155155
Answer: D
Concepts:repeating decimalfactor
Small hint:
A two-digit repeating block means 1n=ab‾99\dfrac{1}{n} = \dfrac{\overline{ab}}{99}
Big hint:
So 99n\dfrac{99}{n} is the two-digit number ab‾;\overline{ab}; check the divisors of 9999
Solution:
If 1n=0.ab‾,\dfrac{1}{n} = 0.\overline{ab}, then 99n=ab‾,\dfrac{99}{n} = \overline{ab}, a two-digit number. The positive divisors of 9999 are 1,3,9,11,33,99.1, 3, 9, 11, 33, 99.
Only n=11,33,99n = 11, 33, 99 make 99n\dfrac{99}{n} equal to 09,03,01,09, 03, 01, which have two different digits. The requested sum is 11+33+99=143.11 + 33 + 99 = 143.
Thus, the correct answer is D.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.