Problem 7 - Entrance Test

We are looking for positive integer solutions (x, y) to the equation 2x + 3y = 100. From the equation, 2x = 100 - 3y. Since 2x is an even number and 100 is an even number, 3y must also be an even number. For 3y to be even, y must be an even integer. Also, x and y must be positive integers, so x >= 1 and y >= 1. Since 2x > 0, we have 100 - 3y > 0 => 3y < 100 => y < 100/3 => y < 33.33... Since y must be an even positive integer, the possible values for y are 2, 4, 6, ..., up to the largest even integer less than 33.33, which is 32. Let y = 2k for some positive integer k. Substitute y = 2k into the equation: 2x + 3(2k) = 100 2x + 6k = 100 Divide by 2: x + 3k = 50 x = 50 - 3k. Since x must be a positive integer, x >= 1: 50 - 3k >= 1 49 >= 3k k <= 49/3 k <= 16.33... Since k must be a positive integer (because y = 2k and y >= 1 implies k >= 1/2), the possible integer values for k are 1, 2, 3, ..., 16. Each value of k corresponds to a unique pair (x, y). For example: If k=1, y=2, x=50-3(1)=47. (2*47 + 3*2 = 94 + 6 = 100) If k=16, y=32, x=50-3(16)=50-48=2. (2*2 + 3*32 = 4 + 96 = 100) There are 16 possible integer values for k, so there are 16 pairs of positive integers (x, y) that satisfy the equation.