2002 AMC 12A Problem 20

Problem 20 of 25HarderNumber Theory

Suppose that

a

and

b

are digits, not both nine and not both zero, and the repeating decimal

0.\overline{ab}

is expressed as a fraction in lowest terms. How many different denominators are possible?

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Solution

Since

0.\overline{ab} = \dfrac{\overline{ab}}{99},

the reduced denominator divides

99 = 3^2\cdot 11.

The divisors are

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

The denominator

1

would require

\overline{ab} = 99,

i.e.

a = b = 9,

which is excluded. Each is achievable: numerators

33, 11, 9, 3,

and

1

reduce to denominators

3, 9, 11, 33,

and

99,

respectively. Thus there are

5

possible denominators. Thus, the correct answer is C.