2021 AMC 12A Spring Problem 5

Attempt Problem 5 of the 2021 AMC 12A Spring 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 2021 AMC 12A Spring solutions, or check the answer key.

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5.

When a student multiplied the number 6666 by the repeating decimal 1.ab=1.ababab, 1.\overline{ab} = 1.ababab\ldots, where aa and bb are digits, he did not notice the notation and just multiplied 6666 by the terminating decimal 1.ab.1.ab. Later he found that his answer was 0.50.5 less than the correct answer.

What is the two-digit integer ab?\overline{ab}?

1515

3030

4545

6060

7575

Answer: E
Concepts:repeating decimallinear equation
Difficulty rating: 1370
Solution:

Let n=abn = \overline{ab} be the two-digit integer. Then 1.ab=1+n991.\overline{ab} = 1 + \dfrac{n}{99} while the terminating value is 1.ab=1+n100.1.ab = 1 + \dfrac{n}{100}. The correct product minus the student's product is 66(n99n100)=66n9900=n150. \begin{aligned} &66\left(\frac{n}{99} - \frac{n}{100}\right) \\ &= 66 \cdot \frac{n}{9900} = \frac{n}{150}. \end{aligned}

Setting n150=0.5\dfrac{n}{150} = 0.5 gives n=75.n = 75.

Thus, the correct answer is E.

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