New Year's Countdown Day 1: One More Month
Consider the equation
where is some positive integer. Given that there are exactly four solutions to this equation, where are positive integers, find the sum of all possible values of
This problem is part of the set New Year's Countdown 2017 .
The answer is 85.
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1 solution
31x + ky = 2017 has four positive integer solution. That also means that there are four different x which serves as the solution for a certain k. Take the smallest and the second smallest x, name them x1 and x2 respectively. We can also name the paired y as y1 and y2 respectively. We get 31x1 + ky1 = 31 x2 + ky2 From here, we can get the fact that 31(x2-x1) is divisible by k. We get that lcm (31,k) divides 31 (x2-x1). As we know that 2017 is not divisible by 31, it's obvious that k is not divisible by 31 either. Which basically means, 31k divides 31(x2-x1). So k divides x2-x1 ( I ) Now we need to examine the nature of x1 + k. Obviously, x1 congruents x1+k in modulo k. So as long as 31(x1+k) is less than 2017, it will also have positive integer y as it's pair, as y = (2017 - 31(x1+k))/k. That means that x1+k is also a solution. ( II ) Based on I and II, we can determine that x2 = x1+k. Using the same method, we get that the solutions are (x1, x1+k, x1+2k, and x1+3k). Now that we know that 31x is less than 2017, we need to consider the positive multiples of 31. From the same logic of ( II ), we can also conclude that if k is too small, x1-k and x1+4k should also be solutions. Obviously, x1 needs to be less or equal k for it to work. We also know that if k is too large, x1+3k will be more than 65. So the possible k should satisfy: min {x1} +3k = 1+3k < 66 and 5k>= x1+4k > 65 Now we have the set k = {14, 15, 16, 17, 18, 19, 20, 21} The rest is trying to find the solution for each k by brute force