Difference between revisions of "2022 AIME II Problems/Problem 5"
Isabelchen (talk | contribs) m (→Solution 1) |
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<cmath>a - b = p_1</cmath> | <cmath>a - b = p_1</cmath> | ||
<cmath>b - c = p_2</cmath> | <cmath>b - c = p_2</cmath> | ||
− | <cmath>a - | + | <cmath>a - c = p_3</cmath> |
− | + | <math>p_3 = a - c = a - b + b - c = p_1 + p_2</math>. Because <math>p_3</math> is the sum of two primes, <math>p_1</math> and <math>p_2</math>, <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_3 = p_2 + 2</math>. There are only <math>8</math> primes less than <math>20</math>: <math>2, 3, 5, 7, 11, 13, 17, 19</math>. Only <math>3, 5, 11, 17</math> plus <math>2</math> equals another prime. <math>p_2 \in \{ 3, 5, 11, 17 \}</math>. | |
− | Once <math>a</math> is determined, <math>b</math> and <math>c</math> | + | Once <math>a</math> is determined, <math>b = a+2</math> and <math>c = b + p_2</math>. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>, and <math>4</math> values of <math>p_2</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{72}}</math> |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Revision as of 04:22, 19 February 2022
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
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.