Difference between revisions of "2020 AMC 10B Problems/Problem 4"
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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, | + | 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== | ==Solution 2== |
Revision as of 23:11, 8 February 2020
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
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
makes the other angle
which is prime, therefore our answer is
Video Solution
~IceMatrix
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.