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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==
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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>
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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>
  
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==Solution==
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==Solution 1==
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It is known that <math>0.\overline{ab}=\frac{ab}{99}</math> and <math>0.ab=\frac{ab}{100}</math>. Let <math>\overline {ab} = x</math>. We have that <math>66(1+\frac{x}{100})+0.5=66(1+\frac{x}{99})</math>. Solving gives that <math>100x-75=99x</math> so <math>x=\boxed{\text{(E)} 75}</math>. ~aop2014
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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
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<cmath>\begin{align*}
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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 \\
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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 \\
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\underline{0}.\overline{\underline{a} \ \underline{b}}&=\frac{25}{33} \\
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\underline{0}.\overline{\underline{a} \ \underline{b}}&=0.\overline{75} \\
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\underline{a} \ \underline{b}&=\boxed{\textbf{(E) }75}.
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\end{align*}</cmath>
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~MRENTHUSIASM
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 +
==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>  
 +
 
 +
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
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 +
==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>
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Expanding both sides, we have <cmath>66 + \frac{33(10a+b)}{50} + \frac{1}{2} = 66 + \frac{2(10a+b)}{3}.</cmath>
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Subtracting <math>66</math> from both sides, we have <cmath>\frac{33(10a+b)}{50} + \frac{1}{2} = \frac{2(10a+b)}{3}.</cmath>
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Multiplying both sides by <math>50 \cdot 3 = 150,</math> we have <cmath>99(10a+b) + 75 = 100(10a+b).</cmath>
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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)
 +
 
 +
==Solution 4 (Answer Choices & Modular) ==
 +
 
 +
Let <math>\underline{a} \ \underline{b}</math> represent the two-digit number <math>ab</math>, not the product of the digits <math>a</math> and <math>b</math>. We can construct fractions for the values <math>1.\underline{a} \ \underline{b}</math>, and <math>1.\overline{\underline{a} \ \underline{b}}</math>, which are <math>\frac{1\underline{a} \ \underline{b}}{100}</math> and <math>\frac{1\underline{a} \ \underline{b}}{99}</math>  respectively. Multiplying by <math>66</math> on both sides and adding <math>1/2</math> to <math>\frac{1\underline{a} \ \underline{b}}{100}</math> and simplifying, we get this: <cmath>\frac{33 * \underline{a}\underline{b} + 25}{50} = \frac{2\underline{a}\underline{b}}{3}.</cmath>
 +
 
 +
Looking at the answer choices, we notice that all of them are divisible by <math>3</math>. This means that since the right-hand side will result in an integer, the left-hand side needs to as well. This means that the numerator of the left-hand side fraction has to be divisible by <math>50</math>. So, we get this expression:
 +
 
 +
<math>33 * \underline{a}\underline{b}\ + 25 \equiv 0\pmod{50}. </math>
 +
 
 +
<math>33 * \underline{a}\underline{b}\ \equiv -25\pmod{50}. </math>
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 +
<math>33 * \underline{a}\underline{b}\ \equiv 25\pmod{50}. </math>
 +
 
 +
This means that the product of <math>33</math> and <math>\underline{a}\underline{b}</math> must have a remainder of <math>25</math> when divided by <math>50</math>. Since it must have a remainder of <math>25</math>, the product should have a units digit of <math>5</math>, which eliminates <math>\textbf{(B) }</math> and <math>\textbf{(D) }</math>. Multiplying <math>33</math> to the rest of the answer choices, the only one which fills this requirement is <math>75</math>, which is <math>\boxed{\textbf{(E) }75}.</math>
 +
 
 +
~neeyakkid23
 +
 
 +
==Video Solution (Simple & Quick)==
 +
https://youtu.be/9HI79V-vtCU
 +
 
 +
~ Education, the Study of Everything
  
 
==Video Solution by Aaron He==
 
==Video Solution by Aaron He==
 
https://www.youtube.com/watch?v=xTGDKBthWsw&t=4m12s
 
https://www.youtube.com/watch?v=xTGDKBthWsw&t=4m12s
  
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==Video Solution(Use of properties of repeating decimals) ==
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==Video Solution (Use of Properties of Repeating Decimals) ==
 
https://www.youtube.com/watch?v=zS1u-ohUDzQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=6\
 
https://www.youtube.com/watch?v=zS1u-ohUDzQ&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=6\
  
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https://www.youtube.com/watch?v=P5al76DxyHY
 
https://www.youtube.com/watch?v=P5al76DxyHY
  
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== Video Solution (Using repeating decimal properties) ==
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== Video Solution by OmegaLearn (Using Repeating Decimal Properties) ==
 
https://youtu.be/vQZ13WiL4WU
 
https://youtu.be/vQZ13WiL4WU
  
 
~ pi_is_3.14
 
~ pi_is_3.14
  
−
==Video Solution, Simple and Quick==
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==Video Solution==
−
https://youtu.be/9HI79V-vtCU
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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
  
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Education, the Study of Everything
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==Video Solution by The Learning Royal==
 +
https://youtu.be/AWjOeBFyeb4
  
 
==See also==
 
==See also==

Latest revision as of 10:18, 28 September 2024

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)

Solution 4 (Answer Choices & Modular)

Let $\underline{a} \ \underline{b}$ represent the two-digit number $ab$, not the product of the digits $a$ and $b$. We can construct fractions for the values $1.\underline{a} \ \underline{b}$, and $1.\overline{\underline{a} \ \underline{b}}$, which are $\frac{1\underline{a} \ \underline{b}}{100}$ and $\frac{1\underline{a} \ \underline{b}}{99}$ respectively. Multiplying by $66$ on both sides and adding $1/2$ to $\frac{1\underline{a} \ \underline{b}}{100}$ and simplifying, we get this: \[\frac{33 * \underline{a}\underline{b} + 25}{50} = \frac{2\underline{a}\underline{b}}{3}.\]

Looking at the answer choices, we notice that all of them are divisible by $3$. This means that since the right-hand side will result in an integer, the left-hand side needs to as well. This means that the numerator of the left-hand side fraction has to be divisible by $50$. So, we get this expression:

$33 * \underline{a}\underline{b}\ + 25 \equiv 0\pmod{50}.$

$33 * \underline{a}\underline{b}\ \equiv -25\pmod{50}.$

$33 * \underline{a}\underline{b}\ \equiv 25\pmod{50}.$

This means that the product of $33$ and $\underline{a}\underline{b}$ must have a remainder of $25$ when divided by $50$. Since it must have a remainder of $25$, the product should have a units digit of $5$, which eliminates $\textbf{(B) }$ and $\textbf{(D) }$. Multiplying $33$ to the rest of the answer choices, the only one which fills this requirement is $75$, which is $\boxed{\textbf{(E) }75}.$

~neeyakkid23

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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