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]]}}
 
{{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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==Video Solution, Simple and Quick==
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https://youtu.be/9HI79V-vtCU
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Education, the Study of Everything
  
 
==Problem==
 
==Problem==

Revision as of 15:16, 15 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.

Video Solution, Simple and Quick

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

Education, the Study of Everything

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 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

See also

2021 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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