2021 JMPSC Invitationals Problems/Problem 3
There are exactly even positive integers less than or equal to that are divisible by . What is the sum of all possible positive integer values of ?
must have exactly 5 even multiples less than . We have two cases, either is odd or even. If is even, then . We solve the inequality to find , but since must be an integer we have x = 18, 20. If is odd, then we can set up the inequality . Solving for the integers must be . The sum is or
Suppose is odd. We have for must work for . Clearly , which means the maximum value that can take on is , and the minimum value it can take on is . Since we need exactly 5 even integers, only will work. Now, suppose is even. We have , which means hold exactly even integer multiples. The answer is
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