Difference between revisions of "2020 AMC 10B Problems/Problem 4"

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{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #4]] and [[2020 AMC 12B Problems|2020 AMC 12B #4]]}}
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==Problem==
 
==Problem==
  
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==Solution==
 
==Solution==
  
Since the three angles of a triangle add up to <math>180^{\circ}</math> and one of the angles is <math>90^{\circ}</math> because it's a right triangle, then <math>a^{\circ} + b^{\circ} = 90^{\circ}</math>.
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Since the three angles of a triangle add up to <math>180^{\circ}</math> and one of the angles is <math>90^{\circ}</math> because it's a right triangle, <math>a^{\circ} + b^{\circ} = 90^{\circ}</math>.
  
 
The greatest prime number less than <math>90</math> is <math>89</math>. If <math>a=89^{\circ}</math>, then <math>b=90^{\circ}-89^{\circ}=1^{\circ}</math>, which is not prime.
 
The greatest prime number less than <math>90</math> is <math>89</math>. If <math>a=89^{\circ}</math>, then <math>b=90^{\circ}-89^{\circ}=1^{\circ}</math>, which is not prime.
  
 
The next greatest prime number less than <math>90</math> is <math>83</math>. If <math>a=83^{\circ}</math>, then <math>b=7^{\circ}</math>, which IS prime, so we have our answer <math>\boxed{\textbf{(D)}\ 7}</math> ~quacker88
 
The next greatest prime number less than <math>90</math> is <math>83</math>. If <math>a=83^{\circ}</math>, then <math>b=7^{\circ}</math>, which IS prime, so we have our answer <math>\boxed{\textbf{(D)}\ 7}</math> ~quacker88
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==Solution 2==
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Looking at the answer choices, only <math>7</math> and <math>11</math> are coprime to <math>90</math>. Testing <math>7</math>, the smaller angle, makes the other angle <math>83</math> which is prime, therefore our answer is <math>\boxed{\textbf{(D)}\ 7}</math>
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==Video Solution==
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https://youtu.be/Gkm5rU5MlOU
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~IceMatrix
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https://youtu.be/y_nsQ7pO63c
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~savannahsolver
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==See Also==
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{{AMC10 box|year=2020|ab=B|num-b=3|num-a=5}}
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{{AMC12 box|year=2020|ab=B|num-b=3|num-a=5}}
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{{MAA Notice}}

Revision as of 12:47, 17 June 2020

The following problem is from both the 2020 AMC 10B #4 and 2020 AMC 12B #4, so both problems redirect to this page.

Problem

The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$, where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$?

$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\  5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11$

Solution

Since the three angles of a triangle add up to $180^{\circ}$ and one of the angles is $90^{\circ}$ because it's a right triangle, $a^{\circ} + b^{\circ} = 90^{\circ}$.

The greatest prime number less than $90$ is $89$. If $a=89^{\circ}$, then $b=90^{\circ}-89^{\circ}=1^{\circ}$, which is not prime.

The next greatest prime number less than $90$ is $83$. If $a=83^{\circ}$, then $b=7^{\circ}$, which IS prime, so we have our answer $\boxed{\textbf{(D)}\ 7}$ ~quacker88

Solution 2

Looking at the answer choices, only $7$ and $11$ are coprime to $90$. Testing $7$, the smaller angle, makes the other angle $83$ which is prime, therefore our answer is $\boxed{\textbf{(D)}\ 7}$

Video Solution

https://youtu.be/Gkm5rU5MlOU

~IceMatrix

https://youtu.be/y_nsQ7pO63c

~savannahsolver

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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
2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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. AMC logo.png