Difference between revisions of "2021 AMC 12A Problems/Problem 5"
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Revision as of 18:56, 26 May 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
- 1 Problem
- 2 Solution 1
- 3 Video Solution, Simple & Quick
- 4 Video Solution by Aaron He
- 5 Video Solution(Use of properties of repeating decimals)
- 6 Video Solution by Punxsutawney Phil
- 7 Video Solution by Hawk Math
- 8 Video Solution (Using repeating decimal properties)
- 9 Video Solution
- 10 Video Solution by TheBeautyofMath
- 11 See also
Problem
When a student multiplied the number by the repeating decimal
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
It is known that and
. Let the 2-digit integer
.
We have that . Expanding and simplifying gives
so
.
~aop2014 ~BakedPotato66
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 (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
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.