2021 AMC 10A Problem 8

Problem 8 of 25EasierAlgebra Number Theory

When a student multiplied the number

66

by the repeating decimal,

\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},

where

a

and

b

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

66

times

\underline{1}.\underline{a} \ \underline{b}.

Later he found that his answer is

0.5

less than the correct answer. What is the

2

-digit integer

\underline{a} \ \underline{b}?

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Solution

Let

N=10a+b,

the two-digit integer formed by the digits

a

and

b.

Then

1.\overline{ab}=1+\frac{N}{99}

while the terminating decimal the student used is

1.ab=1+\frac{N}{100}.

The correct product exceeds the student’s product by

0.5,

66\left(\frac{N}{99}-\frac{N}{100}\right) =\frac{N}{150}=0.5.

Hence

N=75.

Thus, E is the correct answer.