Difference between revisions of "2021 Fall AMC 10B Problems/Problem 3"
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<math>(\textbf{A})\: 1\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 2020\qquad(\textbf{D}) \: 2021\qquad(\textbf{E}) \: 4041</math> | <math>(\textbf{A})\: 1\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 2020\qquad(\textbf{D}) \: 2021\qquad(\textbf{E}) \: 4041</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | We write the given expression as a single fraction: <cmath>\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}</cmath> by cross multiplication. Then by factoring the numerator, we get <cmath>\frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}=\frac{(2021-2020)(2021+2020)}{2020\cdot2021}.</cmath> The question is asking for the numerator, so our answer is <math>2021+2020=4041,</math> giving answer | + | We write the given expression as a single fraction: <cmath>\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}</cmath> by cross multiplication. Then by factoring the numerator, we get <cmath>\frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}=\frac{(2021-2020)(2021+2020)}{2020\cdot2021}.</cmath> The question is asking for the numerator, so our answer is <math>2021+2020=4041,</math> giving <math>\boxed{\textbf{(E) }4041}</math>. |
+ | |||
+ | ~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] | ||
+ | |||
+ | == Solution 2 == | ||
+ | Denote <math>a = 2020</math>. Hence, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{2021}{2020} - \frac{2020}{2021} | ||
+ | & = \frac{a + 1}{a} - \frac{a}{a + 1} \\ | ||
+ | & = \frac{\left( a + 1 \right)^2 - a^2}{a \left( a + 1 \right)} \\ | ||
+ | & = \frac{2 a + 1}{a \left( a + 1 \right)} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | We observe that <math>{\rm gcd} \left( 2a + 1 , a \right) = 1</math> and <math>{\rm gcd} \left( 2a + 1 , a + 1 \right) = 1</math>. | ||
+ | |||
+ | Hence, <math>{\rm gcd} \left( 2a + 1 , a \left( a + 1 \right) \right) = 1</math>. | ||
+ | |||
+ | Therefore, <math>p = 2 a + 1 = 4041</math>. | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(E) }4041}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/p9_RH4s-kBA?t=160 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/ludy6AnQkrI | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/PPIZH_iBTJw | ||
+ | |||
+ | ~savannahsolver | ||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/lC7naDZ1Eu4?t=378 | ||
+ | |||
+ | ~IceMatrix | ||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2021 Fall|ab=B|num-a=4|num-b=2}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:53, 30 December 2022
Contents
Problem
The expression is equal to the fraction
in which
and
are positive integers whose greatest common divisor is
. What is
Solution 1
We write the given expression as a single fraction: by cross multiplication. Then by factoring the numerator, we get
The question is asking for the numerator, so our answer is
giving
.
Solution 2
Denote . Hence,
We observe that and
.
Hence, .
Therefore, .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=160
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/lC7naDZ1Eu4?t=378
~IceMatrix
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.