2021 AMC 12A Spring Problem 5

Question

When a student multiplied the number

66

by the repeating decimal

1.\overline{ab} = 1.ababab\ldots,

where

a

and

b

are digits, he did not notice the notation and just multiplied

66

by the terminating decimal

1.ab.

Later he found that his answer was

0.5

less than the correct answer.

What is the two-digit integer

\overline{ab}?

15

30

45

60

75

Write

n = \overline{ab}.

Then

1.\overline{ab} = 1 + \dfrac{n}{99}

and

1.ab = 1 + \dfrac{n}{100}

The difference of the two products is

66\left(\dfrac{n}{99} - \dfrac{n}{100}\right) = 0.5

Let

n = \overline{ab}

be the two-digit integer. Then

1.\overline{ab} = 1 + \dfrac{n}{99}

while the terminating value is

1.ab = 1 + \dfrac{n}{100}.

The correct product minus the student’s product is

\begin{aligned} &66\left(\frac{n}{99} - \frac{n}{100}\right) \\ &= 66 \cdot \frac{n}{9900} = \frac{n}{150}. \end{aligned}

Setting

\dfrac{n}{150} = 0.5

gives

n = 75.

Thus, the correct answer is E.

Problem 5 in other years