Difference between revisions of "2021 AMC 12A Problems/Problem 5"
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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 | 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 | ||
| + | ==Video Solution by Aaron He== | ||
| + | https://www.youtube.com/watch?v=xTGDKBthWsw&t=4m12s | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
https://youtube.com/watch?v=MUHja8TpKGw&t=359s | https://youtube.com/watch?v=MUHja8TpKGw&t=359s | ||
Revision as of 22:20, 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.
Contents
Problem
When a student multiplied the number 
by the repeating decimal ![\[\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},\]](http://latex.artofproblemsolving.com/6/0/a/60a48ebc868080ee1035b06da5cf73954173c68c.png)
where 
and 
are digits, he did not notice the notation and just multiplied times 
. Later he found that his answer is 
less than the correct answer. What is the 
-digit number 
Solution
It is known that 
and 
. Let 
. We have that 
. Solving gives that 
so 
. ~aop2014
Video Solution by Aaron He
https://www.youtube.com/watch?v=xTGDKBthWsw&t=4m12s
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)
~ 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. 
