2020 AMC 10B Problems/Problem 4

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

https://youtu.be/Gkm5rU5MlOU

~IceMatrix

https://youtu.be/y_nsQ7pO63c

~savannahsolver

https://youtu.be/wH7xhYxwaFc

~AlexExplains

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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