XVI OM - I - Problem 4

The school organized three trips for its 300 students. The same number of students participated in each trip. Each student went on at least one trip, but half of the participants in the first trip, one-third of the participants in the second trip, and one-fourth of the participants in the third trip only went on one trip.
How many students went on each trip? How many participants in the first trip also participated in the second, and how many of them also participated in the third trip?
Let $ x $ denote the number of participants in each trip, and $ y $, $ z $, $ u $, $ w $ denote, respectively, the numbers of students who went: a) only on the first and second trip, b) only on the first and third trip, c) only on the second and third trip, d) on all three trips.
The total number of students equals the sum of: 1) the number of participants in only one trip, i.e., the number $ \frac{x}{2} + \frac{x}{3} + \frac{x}{4} = \frac{13}{12}x $, 2) the number of participants in only two trips, and 3) the number of participants in all three trips, thus



The number of participants in the first trip is the sum of the numbers: 1) students who went only on the first trip, 2) students who went on the first and second or first and third trip, and 3) students who went on all three trips, thus






Similarly, for the number of participants in the second and third trips, we obtain the equations












We need to find non-negative integer solutions to the system of equations (1) - (4). Suppose such a solution is the set of numbers $ x $, $ y $, $ z $, $ u $, $ w $. From equation (1), it follows that $ \frac{13}{12} x $ is an integer, so $ x $ is divisible by $ 12 $. Let $ x = 12t $; equations (1) - (4) take the form
























Adding equations (5) and (8) side by side, we get the equation






Since $ y \geq 0 $, it follows from (9) that $ t \leq \frac{300}{22}=13.6\ldots $, thus






Adding (6) and (7) side by side and subtracting (8), we get






Since $ w \geq 0 $, it follows from equation (11) that $ y \leq \frac{5}{2} t $, from which, using equation (9), we conclude that $ 22t + \frac{5}{2}t \geq 300 $, thus
$ t \geq \frac{600}{49} = 12.2\ldots, $ hence






From (10) and (11), it follows that






From equations (9), (11), (6), and (7), we obtain sequentially $ y = 14 $, $ w = 37 $, $ z = 27 $, $ u = 53 $. It is easy to verify that the found numbers satisfy the system of equations (1) - (4), thus they constitute the only solution to the problem.