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]]}} | ||
+ | |||
==Problem== | ==Problem== | ||
− | + | When a student multiplied the number <math>66</math> by the repeating decimal, | |
− | ==Solution== | + | <cmath>\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},</cmath> |
− | + | where <math>a</math> and <math>b</math> are digits, he did not notice the notation and just multiplied <math>66</math> times <math>\underline{1}.\underline{a} \ \underline{b}.</math> Later he found that his answer is <math>0.5</math> less than the correct answer. What is the <math>2</math>-digit number <math>\underline{a} \ \underline{b}?</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== | ||
+ | It is known that <math>\underline{0}.\overline{\underline{a} \ \underline{b}}=\frac{\underline{a} \ \underline{b}}{99}</math> and <math>\underline{0}.\underline{a} \ \underline{b}=\frac{\underline{a} \ \underline{b}}{100}.</math> | ||
+ | |||
+ | Let <math>x=\underline{a} \ \underline{b}.</math> We have <cmath>66\biggl(1+\frac{x}{99}\biggr)-66\biggl(1+\frac{x}{100}\biggr)=0.5.</cmath> Expanding and simplifying give <math>\frac{x}{150}=0.5,</math> so <math>x=\boxed{\textbf{(E) }75}.</math> | ||
+ | |||
+ | ~aop2014 ~BakedPotato66 ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 3 (Similar to Solution 2)== | ||
+ | |||
+ | We have <cmath>66 \cdot \left(1 + \frac{10a+b}{100}\right) + \frac{1}{2} = 66 \cdot \left(1+ \frac{10a+b}{99}\right).</cmath> | ||
+ | Expanding both sides, we have <cmath>66 + \frac{33(10a+b)}{50} + \frac{1}{2} = 66 + \frac{2(10a+b)}{3}.</cmath> | ||
+ | Subtracting <math>66</math> from both sides, we have <cmath>\frac{33(10a+b)}{50} + \frac{1}{2} = \frac{2(10a+b)}{3}.</cmath> | ||
+ | Multiplying both sides by <math>50 \cdot 3 = 150,</math> we have <cmath>99(10a+b) + 75 = 100(10a+b).</cmath> | ||
+ | Thus, the answer is <math>10a+b = \boxed{\textbf{(E) }75}.</math> | ||
+ | |||
+ | By letting <math>x=\underline{a} \ \underline{b}=10a+b,</math> this solution is similar to Solution 2. In this solution, we solve for <math>10a+b</math> as a whole. | ||
+ | |||
+ | -mathboy282 (Solution) | ||
+ | |||
+ | ~MRENTHUSIASM (Minor Revision) | ||
+ | |||
+ | ==Video Solution (Simple & Quick)== | ||
+ | https://youtu.be/9HI79V-vtCU | ||
+ | |||
+ | ~ 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) == | ||
+ | https://youtu.be/vQZ13WiL4WU | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/DOF3FYUsXsU | ||
+ | |||
+ | ~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== | ||
+ | https://youtu.be/AWjOeBFyeb4 | ||
+ | |||
==See also== | ==See also== | ||
+ | {{AMC10 box|year=2021|ab=A|num-b=7|num-a=9}} | ||
{{AMC12 box|year=2021|ab=A|num-b=4|num-a=6}} | {{AMC12 box|year=2021|ab=A|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:46, 28 October 2022
- 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,
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
Solution 3 (Similar to Solution 2)
We have
Expanding both sides, we have
Subtracting
from both sides, we have
Multiplying both sides by
we have
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.