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

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It is known that <math>\underline{0}.\overline{\underline{a} \ \underline{b}}=\frac{\underline{a} \ \underline{b}}{99}</math> and <math>\underline{0}.\underline{a} \ \underline{b}=\frac{\underline{a} \ \underline{b}}{100}.</math>  
 
It is known that <math>\underline{0}.\overline{\underline{a} \ \underline{b}}=\frac{\underline{a} \ \underline{b}}{99}</math> and <math>\underline{0}.\underline{a} \ \underline{b}=\frac{\underline{a} \ \underline{b}}{100}.</math>  
  
−
Let <math>x=\underline{a} \ \underline{b}.</math> We have <cmath>66\biggl(1+\frac{x}{99}\biggr)=66\biggl(1+\frac{x}{100}\biggr)+0.5.</cmath> Expanding and simplifying give <math>\frac{x}{150}-0.5,</math> so <math>x=\boxed{\textbf{(E) }75}.</math>  
+
Let <math>x=\underline{a} \ \underline{b}.</math> We have <cmath>66\biggl(1+\frac{x}{99}\biggr)=66\biggl(1+\frac{x}{100}\biggr)+0.5.</cmath> Expanding and simplifying give <math>\frac{x}{150}=0.5,</math> so <math>x=\boxed{\textbf{(E) }75}.</math>  
  
 
~aop2014 ~BakedPotato66 ~MRENTHUSIASM
 
~aop2014 ~BakedPotato66 ~MRENTHUSIASM

Revision as of 17:41, 12 July 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, $\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},$ 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 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

We are given that $66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-0.5=66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr),$ from which \begin{align*} 66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}} - \underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl(\underline{0}.\underline{0} \ \underline{0} \ \overline{\underline{a} \ \underline{b}}\Bigr)&=0.5 \\ 66\left(\frac{1}{100}\cdot\underline{0}.\overline{\underline{a} \ \underline{b}}\right)&=\frac12 \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=\frac{25}{33} \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=0.\overline{75} \\ \underline{a} \ \underline{b}&=\boxed{\textbf{(E) }75}. \end{align*} ~MRENTHUSIASM

Solution 2

It is known that $\underline{0}.\overline{\underline{a} \ \underline{b}}=\frac{\underline{a} \ \underline{b}}{99}$ and $\underline{0}.\underline{a} \ \underline{b}=\frac{\underline{a} \ \underline{b}}{100}.$

Let $x=\underline{a} \ \underline{b}.$ We have \[66\biggl(1+\frac{x}{99}\biggr)=66\biggl(1+\frac{x}{100}\biggr)+0.5.\] Expanding and simplifying give $\frac{x}{150}=0.5,$ so $x=\boxed{\textbf{(E) }75}.$

~aop2014 ~BakedPotato66 ~MRENTHUSIASM

Video Solution (Simple & Quick)

https://youtu.be/9HI79V-vtCU

~ Education, the Study of Everything

Video Solution by Aaron He

https://www.youtube.com/watch?v=xTGDKBthWsw&t=4m12s

Video Solution (Use of Properties of Repeating Decimals)

https://www.youtube.com/watch?v=zS1u-ohUDzQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=6\

~North America Math Contest Go Go Go

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

Video Solution

https://youtu.be/DOF3FYUsXsU

~savannahsolver

Video Solution by TheBeautyofMath

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

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

~IceMatrix

Video Solution by The Learning Royal

https://youtu.be/AWjOeBFyeb4

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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