2021 AMC 12A Problems/Problem 5
- The following problem is from both the 2021 AMC 10A #8 and 2021 AMC 12A #5, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Similar to Solution 2)
- 5 Video Solution (Simple & Quick)
- 6 Video Solution by Aaron He
- 7 Video Solution (Use of Properties of Repeating Decimals)
- 8 Video Solution by Punxsutawney Phil
- 9 Video Solution by Hawk Math
- 10 Video Solution by OmegaLearn (Using Repeating Decimal Properties)
- 11 Video Solution
- 12 Video Solution by TheBeautyofMath
- 13 Video Solution by The Learning Royal
- 14 See also
Problem
When a student multiplied the number 
by the repeating decimal,
![\[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\]](http://latex.artofproblemsolving.com/d/8/a/d8a2b2bed7d27aa171aee4df068bd644bdcb9bb7.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 1
We are given that 
from which
~MRENTHUSIASM
Solution 2
It is known that 
and 
Let 
We have ![\[66\biggl(1+\frac{x}{99}\biggr)-66\biggl(1+\frac{x}{100}\biggr)=0.5.\]](http://latex.artofproblemsolving.com/a/f/8/af80192878db4177e60c0eadda01c922885104a7.png)
Expanding and simplifying give 
so 
~aop2014 ~BakedPotato66 ~MRENTHUSIASM
Solution 3 (Similar to Solution 2)
We have ![\[66 \cdot \left(1 + \frac{10a+b}{100}\right) + \frac{1}{2} = 66 \cdot \left(1+ \frac{10a+b}{99}\right).\]](http://latex.artofproblemsolving.com/3/e/1/3e1444974ba690125cd937fc082d942d2d04a694.png)
Expanding both sides, we have ![\[66 + \frac{33(10a+b)}{50} + \frac{1}{2} = 66 + \frac{2(10a+b)}{3}.\]](http://latex.artofproblemsolving.com/c/7/f/c7f73507badd69f6810af79f1a7b4d67fb1baedb.png)
Subtracting from both sides, we have ![\[\frac{33(10a+b)}{50} + \frac{1}{2} = \frac{2(10a+b)}{3}.\]](http://latex.artofproblemsolving.com/1/2/9/1296f65bfb93bd23720a703ce8f88d197d458812.png)
Multiplying both sides by 
we have ![\[99(10a+b) + 75 = 100(10a+b).\]](http://latex.artofproblemsolving.com/4/5/2/452ae660eaf715b5d0f70fb4d40b50500bf470db.png)
Thus, the answer is 
By letting 
this solution is similar to Solution 2. In this solution, we solve for 
as a whole.
-mathboy282 (Solution)
~MRENTHUSIASM (Minor Revision)
Video Solution (Simple & Quick)
~ 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)
~ pi_is_3.14
Video Solution
~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
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. 
