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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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<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\  5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11</math>
 
<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\  5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11</math>
  
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==Solution==
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==Solution 1==
  
 
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>.
 
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>.
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==Solution 2==
 
==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> makes the other angle <math>83</math> which is prime, therefore our answer is <math>\boxed{\textbf{(D)}\ 7}</math>
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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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==Solution 3 (Euclidean Algorithm)==
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It is clear that <math>\gcd(a,b)=1.</math> By the Euclidean Algorithm, we have <cmath>\gcd(a,b)=\gcd(a+b,b)=\gcd(90,b)=1,</cmath> so <math>90</math> and <math>b</math> are relatively prime.
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The least such prime number <math>b</math> is <math>7,</math> from which <math>a=90-b=83</math> is also a prime number. Therefore, the answer is <math>\boxed{\textbf{(D)}\ 7}.</math>
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~MRENTHUSIASM
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==Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)==
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https://youtu.be/hE2fEOfviDw
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~Education, the Study of Everything
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==Video Solutions==
 
https://youtu.be/Gkm5rU5MlOU
 
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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https://youtu.be/wH7xhYxwaFc
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~AlexExplains
  
 
==See Also==
 
==See Also==

Latest revision as of 12:33, 6 June 2023

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 1

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 $\boxed{\textbf{(D)}\ 7}$

Solution 3 (Euclidean Algorithm)

It is clear that $\gcd(a,b)=1.$ By the Euclidean Algorithm, we have \[\gcd(a,b)=\gcd(a+b,b)=\gcd(90,b)=1,\] so $90$ and $b$ are relatively prime.

The least such prime number $b$ is $7,$ from which $a=90-b=83$ is also a prime number. Therefore, the answer is $\boxed{\textbf{(D)}\ 7}.$

~MRENTHUSIASM

Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)

https://youtu.be/hE2fEOfviDw

~Education, the Study of Everything


Video Solutions

https://youtu.be/Gkm5rU5MlOU


https://youtu.be/y_nsQ7pO63c

~savannahsolver

https://youtu.be/wH7xhYxwaFc

~AlexExplains

See Also

2020 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

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