2013 AMC 12A Problem 10

Problem 10 of 25EasierNumber Theory

Let

S

be the set of positive integers

n

for which

\dfrac{1}{n}

has the repeating decimal representation

0.\overline{ab} = 0.ababab\ldots,

with

a

and

b

different digits. What is the sum of the elements of

S?

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Solution

\dfrac{1}{n} = 0.\overline{ab},

then

\dfrac{99}{n} = \overline{ab},

a two-digit number. The positive divisors of

99

are

1, 3, 9, 11, 33, 99.

Only

n = 11, 33, 99

make

\dfrac{99}{n}

equal to

09, 03, 01,

which have two different digits. The requested sum is

11 + 33 + 99 = 143.

Thus, the correct answer is D.