Difference between revisions of "2021 AMC 12A Problems/Problem 5"
MRENTHUSIASM (talk | contribs) m (→Video Solution (Use of properties of repeating decimals): Capitalized title.) |
MRENTHUSIASM (talk | contribs) (Added in Sol 1, as that's the easiest. I pushed that to top) |
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<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> | ||
| + | |||
| + | ==Solution 1== | ||
| + | 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 | ||
| + | <cmath>\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*}</cmath> | ||
| + | ~MRENTHUSIASM | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 02:22, 4 July 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 Solution 2
- 4 Video Solution (Simple & Quick)
- 5 Video Solution by Aaron He
- 6 Video Solution (Use of Properties of Repeating Decimals)
- 7 Video Solution by Punxsutawney Phil
- 8 Video Solution by Hawk Math
- 9 Video Solution (Using Repeating Decimal Properties)
- 10 Video Solution
- 11 Video Solution by TheBeautyofMath
- 12 Video Solution by The Learning Royal
- 13 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
We are given that 
from which
~MRENTHUSIASM
Solution 2
It is known that 
and 
Let 
We have 
Expanding and simplifying give 
so 
~aop2014 ~BakedPotato66 ~MRENTHUSIASM
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
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. 
