Difference between revisions of "2022 AIME II Problems/Problem 5"
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+ | ===Note: This solution seems incorrect.=== | ||
+ | Although the answer is correct, solution 2 below is a more accurate way to approach this problem. I agree, I don't get how <math>a + 2 \leq 20</math>. | ||
==Solution 2== | ==Solution 2== | ||
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If the primes are <math>2,5,7</math>, then the smallest number can range between <math>1</math> and <math>13</math>. | If the primes are <math>2,5,7</math>, then the smallest number can range between <math>1</math> and <math>13</math>. | ||
If the primes are <math>2,11,13</math>, then the smallest number can range between <math>1</math> and <math>7</math>. | If the primes are <math>2,11,13</math>, then the smallest number can range between <math>1</math> and <math>7</math>. | ||
− | If the primes are <math>2,17,19</math>, then the smallest | + | If the primes are <math>2,17,19</math>, then the smallest number can only be <math>1</math>. |
Adding all cases gets <math>15+13+7+1=36</math>. However, due to the commutative property, we must multiply this by 2. For example, in the <math>2,17,19</math> case the numbers can be <math>1,3,20</math> or <math>1,18,20</math>. Therefore the answer is <math>36\cdot2=\boxed{072}</math>. | Adding all cases gets <math>15+13+7+1=36</math>. However, due to the commutative property, we must multiply this by 2. For example, in the <math>2,17,19</math> case the numbers can be <math>1,3,20</math> or <math>1,18,20</math>. Therefore the answer is <math>36\cdot2=\boxed{072}</math>. | ||
Note about solution 1: I don't think that works, because if for example there are 21 points on the circle, your solution would yield <math>19\cdot4=76</math>, but there would be <math>8</math> more solutions than if there are <math>20</math> points. This is because the upper bound for each case increases by <math>1</math>, but commutative property doubles it to be <math>4</math>. | Note about solution 1: I don't think that works, because if for example there are 21 points on the circle, your solution would yield <math>19\cdot4=76</math>, but there would be <math>8</math> more solutions than if there are <math>20</math> points. This is because the upper bound for each case increases by <math>1</math>, but commutative property doubles it to be <math>4</math>. | ||
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+ | ==Video Solution by Power of Logic== | ||
+ | https://youtu.be/iI2ZpdpGNyc | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=II|num-b=4|num-a=6}} | {{AIME box|year=2022|n=II|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:08, 28 January 2024
Contents
Problem
Twenty distinct points are marked on a circle and labeled through in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original points.
Solution 1
Let , , and be the vertex of a triangle that satisfies this problem, where .
. Because is the sum of two primes, and , or must be . Let , then . There are only primes less than : . Only plus equals another prime. .
Once is determined, and . There are values of where , and values of . Therefore the answer is
Note: This solution seems incorrect.
Although the answer is correct, solution 2 below is a more accurate way to approach this problem. I agree, I don't get how .
Solution 2
As above, we must deduce that the sum of two primes must be equal to the third prime. Then, we can finish the solution using casework. If the primes are , then the smallest number can range between and . If the primes are , then the smallest number can range between and . If the primes are , then the smallest number can range between and . If the primes are , then the smallest number can only be .
Adding all cases gets . However, due to the commutative property, we must multiply this by 2. For example, in the case the numbers can be or . Therefore the answer is .
Note about solution 1: I don't think that works, because if for example there are 21 points on the circle, your solution would yield , but there would be more solutions than if there are points. This is because the upper bound for each case increases by , but commutative property doubles it to be .
Video Solution by Power of Logic
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.