2022 AIME II Problems/Problem 5
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.
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
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 prime 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 .
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