Difference between revisions of "2020 AMC 10A Problems/Problem 3"
Atihsim123 (talk | contribs) m (→Solution 2) |
MRENTHUSIASM (talk | contribs) (Ordered the video solutions. Also, I fixed the coding in Solution 3 so that the -1's are on top of the factors instead of appearing in the same row as the numerator. Relatively, this version is clearer.) |
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== Solution 1 == | == Solution 1 == | ||
− | Note that <math>a-3</math> is <math>-1</math> times <math>3-a</math>. Likewise, <math>b-4</math> is <math>-1</math> times <math>4-b</math> and <math>c-5</math> is <math>-1</math> times <math>5-c</math>. Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A)}-1}</math>. | + | Note that <math>a-3</math> is <math>-1</math> times <math>3-a</math>. Likewise, <math>b-4</math> is <math>-1</math> times <math>4-b</math> and <math>c-5</math> is <math>-1</math> times <math>5-c</math>. Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}</math>. |
== Solution 2 == | == Solution 2 == | ||
Substituting values for <math>a, b,\text{ and } c</math>, we see that if each of them satisfy the inequalities above, the value goes to be <math>-1</math>. | Substituting values for <math>a, b,\text{ and } c</math>, we see that if each of them satisfy the inequalities above, the value goes to be <math>-1</math>. | ||
− | Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A)}-1}</math>. | + | Therefore, the product of the given fraction equals <math>(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}</math>. |
== Solution 3 == | == Solution 3 == | ||
− | It is known that <math>\frac{x-y}{y-x}=-1</math> for <math>x\ne y</math>. We use this fact to cancel out the terms. | + | It is known that <math>\frac{x-y}{y-x}=-1</math> for <math>x\ne y</math>. We use this fact to cancel out the terms: <cmath>\frac{\overset{-1}{\cancel{a-3}}}{\underset{1}{\xcancel{5-c}}} \cdot \frac{\overset{-1}{\bcancel{b-4}}}{\underset{1}{\cancel{3-a}}} \cdot \frac{\overset{-1}{\xcancel{c-5}}}{\underset{1}{\bcancel{4-b}}}=(-1)(-1)(-1)=\boxed{\textbf{(A) } -1}.</cmath> |
− | <math>\ | + | ~CoolJupiter (Solution) |
− | + | ~MRENTHUSIASM (<math>\LaTeX</math> Adjustment) | |
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== Video Solution 1 == | == Video Solution 1 == | ||
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~savannahsolver | ~savannahsolver | ||
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+ | == Video Solution 5== | ||
+ | https://youtu.be/ba6w1OhXqOQ?t=956 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2020|ab=A|num-b=2|num-a=4}} | {{AMC10 box|year=2020|ab=A|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:16, 17 April 2021
Contents
Problem
Assuming ,
, and
, what is the value in simplest form of the following expression?
Solution 1
Note that is
times
. Likewise,
is
times
and
is
times
. Therefore, the product of the given fraction equals
.
Solution 2
Substituting values for , we see that if each of them satisfy the inequalities above, the value goes to be
.
Therefore, the product of the given fraction equals
.
Solution 3
It is known that for
. We use this fact to cancel out the terms:
~CoolJupiter (Solution)
~MRENTHUSIASM ( Adjustment)
Video Solution 1
~IceMatrix
Video Solution 2
Education, The Study of Everything
Video Solution 3
https://www.youtube.com/watch?v=7-3sl1pSojc
~bobthefam
Video Solution 4
~savannahsolver
Video Solution 5
https://youtu.be/ba6w1OhXqOQ?t=956
~ pi_is_3.14
See Also
2020 AMC 10A (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.