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Difference between revisions of "2021 AMC 12A Problems/Problem 5"

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{{duplicate|[[2021 AMC 10A Problems#Problem 8|2021 AMC 10A #8]] and [[2021 AMC 12A Problems#Problem 5|2021 AMC 12A #5]]}}
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==Problem==
 
==Problem==
 
When a student multiplied the number <math>66</math> by the repeating decimal <cmath>\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},</cmath> where <math>a</math> and <math>b</math> are digits, he did not notice the notation and just multiplied <math>66</math> times <math>\underline{1}.\underline{a}\underline{b}</math>. Later he found that his answer is <math>0.5</math> less than the correct answer. What is the <math>2</math>-digit number <math>\underline{a}\underline{b}?</math>
 
When a student multiplied the number <math>66</math> by the repeating decimal <cmath>\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},</cmath> where <math>a</math> and <math>b</math> are digits, he did not notice the notation and just multiplied <math>66</math> times <math>\underline{1}.\underline{a}\underline{b}</math>. Later he found that his answer is <math>0.5</math> less than the correct answer. What is the <math>2</math>-digit number <math>\underline{a}\underline{b}?</math>

Revision as of 17:37, 12 February 2021

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

It is known that $0.\overline{ab}=\frac{ab}{99}$ and $0.ab=\frac{ab}{100}$. Let $\overline {ab} = x$. We have that $66(1+\frac{x}{100})+0.5=66(1+\frac{x}{99})$. Solving gives that $100x-75=99x$ so $x=\boxed{\text{(E)} 75}$. ~aop2014

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=MUHja8TpKGw&t=359s

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

Video Solution (Using repeating decimal properties)

https://youtu.be/vQZ13WiL4WU

~ pi_is_3.14

See also

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

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