Difference between revisions of "2020 AMC 10B Problems/Problem 4"
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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> | ||
− | ==Solution== | + | ==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>. |
Revision as of 19:15, 1 May 2021
- The following problem is from both the 2020 AMC 10B #4 and 2020 AMC 12B #4, so both problems redirect to this page.
Contents
[hide]Problem
The acute angles of a right triangle are and , where and both and are prime numbers. What is the least possible value of ?
Solution 1
Since the three angles of a triangle add up to and one of the angles is because it's a right triangle, .
The greatest prime number less than is . If , then , which is not prime.
The next greatest prime number less than is . If , then , which IS prime, so we have our answer ~quacker88
Solution 2
Looking at the answer choices, only and are coprime to . Testing , the smaller angle, makes the other angle which is prime, therefore our answer is
Video Solution
~IceMatrix
~savannahsolver
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.