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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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{{duplicate|[[2021 AMC 10A Problems/Problem 8|2021 AMC 10A #8]] and [[2021 AMC 12A Problems/Problem 5|2021 AMC 12A #5]]}}
  
 
==Problem==
 
==Problem==
−
When a student multiplied the number <math>66</math> by the repeating decimal, <math>\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},</math> 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}\ldots=\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>
  
 
<math>\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75</math>
 
<math>\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75</math>
 +
 +
==Solution 1==
 +
We are given that <math>66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-0.5=66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr),</math> from which
 +
<cmath>\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*}</cmath>
 +
~MRENTHUSIASM
  
 
==Solution 2==
 
==Solution 2==
 
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 <math>66\left(1+\frac{x}{100}\right)+0.5=66\left(1+\frac{x}{99}\right).</math> Expanding and simplifying give <math>0.5=\frac{x}{150},</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
 +
 +
==Solution 3 (Similar to Solution 2)==
 +
 +
We have <cmath>66 \cdot \left(1 + \frac{10a+b}{100}\right) + \frac{1}{2} = 66 \cdot \left(1+ \frac{10a+b}{99}\right).</cmath>
 +
Expanding both sides, we have <cmath>66 + \frac{33(10a+b)}{50} + \frac{1}{2} = 66 + \frac{2(10a+b)}{3}.</cmath>
 +
Subtracting <math>66</math> from both sides, we have <cmath>\frac{33(10a+b)}{50} + \frac{1}{2} = \frac{2(10a+b)}{3}.</cmath>
 +
Multiplying both sides by <math>50 \cdot 3 = 150,</math> we have <cmath>99(10a+b) + 75 = 100(10a+b).</cmath>
 +
Thus, the answer is <math>10a+b = \boxed{\textbf{(E) }75}.</math>
 +
 +
By letting <math>x=\underline{a} \ \underline{b}=10a+b,</math> this solution is similar to Solution 2. In this solution, we solve for <math>10a+b</math> as a whole.
 +
 +
-mathboy282 (Solution)
 +
 +
~MRENTHUSIASM (Minor Revision)
  
 
==Video Solution (Simple & Quick)==
 
==Video Solution (Simple & Quick)==
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https://www.youtube.com/watch?v=P5al76DxyHY
 
https://www.youtube.com/watch?v=P5al76DxyHY
  
−
== Video Solution (Using Repeating Decimal Properties) ==
+
== Video Solution by OmegaLearn (Using Repeating Decimal Properties) ==
 
https://youtu.be/vQZ13WiL4WU
 
https://youtu.be/vQZ13WiL4WU
  

Revision as of 21:46, 28 October 2022

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

Solution 3 (Similar to Solution 2)

We have \[66 \cdot \left(1 + \frac{10a+b}{100}\right) + \frac{1}{2} = 66 \cdot \left(1+ \frac{10a+b}{99}\right).\] Expanding both sides, we have \[66 + \frac{33(10a+b)}{50} + \frac{1}{2} = 66 + \frac{2(10a+b)}{3}.\] Subtracting $66$ from both sides, we have \[\frac{33(10a+b)}{50} + \frac{1}{2} = \frac{2(10a+b)}{3}.\] Multiplying both sides by $50 \cdot 3 = 150,$ we have \[99(10a+b) + 75 = 100(10a+b).\] Thus, the answer is $10a+b = \boxed{\textbf{(E) }75}.$

By letting $x=\underline{a} \ \underline{b}=10a+b,$ this solution is similar to Solution 2. In this solution, we solve for $10a+b$ as a whole.

-mathboy282 (Solution)

~MRENTHUSIASM (Minor Revision)

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 by OmegaLearn (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

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