Difference between revisions of "2020 AMC 10B Problems/Problem 4"
Kevinmathz (talk | contribs) (→Solution 2) |
(→Video Solutions) |
||
| (10 intermediate revisions by 5 users not shown) | |||
| Line 7: | Line 7: | ||
<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>. | ||
| Line 19: | Line 19: | ||
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> | 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> | ||
| − | ==Video Solution== | + | ==Solution 3 (Euclidean Algorithm)== |
| + | 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. | ||
| + | |||
| + | 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> | ||
| + | |||
| + | ~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/Gkm5rU5MlOU | ||
| − | ~ | + | |
| + | https://youtu.be/y_nsQ7pO63c | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | https://youtu.be/wH7xhYxwaFc | ||
| + | |||
| + | ~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.
Contents
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
Solution 3 (Euclidean Algorithm)
It is clear that 
By the Euclidean Algorithm, we have ![\[\gcd(a,b)=\gcd(a+b,b)=\gcd(90,b)=1,\]](http://latex.artofproblemsolving.com/2/e/c/2ecd80a8a70cb0f3bc800f0ddc3d93571157b77a.png)
so and are relatively prime.
The least such prime number is 
from which 
is also a prime number. Therefore, the answer is 
~MRENTHUSIASM
Video Solution (HOW TO CREATIVELY PROBLEM SOLVE!!!)
~Education, the Study of Everything
Video Solutions
~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. 
