# 2021 AMC 12A Problems/Problem 5

The following problem is from both the 2021 AMC 10A #8 and 2021 AMC 12A #5, so both problems redirect to this page.

## Problem

When a student multiplied the number $66$ by the repeating decimal $$\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},$$ 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 number $\underline{a}\underline{b}?$ $\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$

## Solution 1

It is known that $0.\overline{ab}=\frac{ab}{99}$ and $0.ab=\frac{ab}{100}$. Let the 2-digit integer $\underline {ab} = x$.

We have that $66(1+\frac{x}{100})+0.5=66(1+\frac{x}{99})$. Expanding and simplifying gives $0.5=\dfrac{x}{150}$ so $x=\boxed{\text{(E)} 75}$.

~aop2014 ~BakedPotato66

## Video Solution, Simple & Quick

~ Education, the Study of Everything

## Video Solution(Use of properties of repeating decimals)

~North America Math Contest Go Go Go

~ pi_is_3.14

~savannahsolver

## Video Solution by TheBeautyofMath

https://youtu.be/s6E4E06XhPU?t=360 (AMC 10A)

https://youtu.be/rEWS75W0Q54?t=511 (AMC 12A)

~IceMatrix

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 