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
Difficulty rating: 1510
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.

← Problem 9#9
Full Exam

Problem 10 in Other Years