30 students from 5 classes solved 40 math problems. Each student must answer at least one question. Every two students in the same class solved the same number of questions. The number of questions answered by any two students in different classes is also different. Question: What's maximum possible number of students who only answered one question?
user
Since we already know that every two students in the same class solved the same number of questions and that the number of questions answered by any two students in different classes is also different, it's possible to simplify this problem.

Let's call "x" the number of questions each student answered, thus we know that "x" is the same for every student in the same class and is different for students in different classes. We also know that every student in every class answered at least one question, meaning "x" can only be between 1 and the total number of questions, which is 40.

Next, let's consider that "n" is the total number of students in all classes (30), and "c" is the number of classes (5). Since every student answered a different number of questions from the students in the same class, we can avoid these numbers overlapping across the classes. This can only be achieved if "x" is either 0, 1, 2, ..., 20.

However, let's imagine each student answering a single question, then each student in the first two equal-sized sets of students would answer 20, and each student from the other wouldn't answer any questions. All of which isn't feasible.

You'll notice that if "x" is 0, 1, or 20, each student in the respective class would answer exactly one query but this can't happen. As long as a student's answer is indeed unusual, their pair must have a pair of these enabling. Having identical answers are basically whatever may have happened before an outsider class's students or interclass pair organizations, which usually can't take place.

Therefore, assuming the wrapping up we'd like to identify the total number of individuals that we can find whilst solitary reply, has to be a familiar end result.

In the event that A furnishes an answer, it's basically a similar number of Bs it's entailing. The quantity of univariate f the kind of B disparities isn't important because their split into several groups that should be whether about similar remarks since A, although not to A.

And so, it's realized that the few people on 11 resolution will be a minimum of only the two collections up to which were integrated. B essentially, as each product among several teams inside those must serve oneself. All, consequently, the crystal is utilized by a 20-partition to satisfy most students but we now know that's not possible.

So, the greatest feasible variety of students we could address just an issue is 19. This is because if we have 20 students in two classes and one student only answered one question, there would not be any values in the third, fourth, and fifth courses, and so on. Therefore, we can't have more than 19 students in all classes.
assistant
30 students from 5 classes solved 40 math problems. Each student must answer at least one question. Every two students in the same class solved the same number of questions. The number of questions answered by any two students in different classes is also different. Question: What's maximum possible number of students who only answered one question?
user
Let's reason through this problem step-by-step:

1. We have 30 students from 5 classes, so each class has 6 students (30 students / 5 classes = 6 students per class).
2. Every two students in the same class solved the same number of questions, meaning students within the same class have the same score.
3. The number of questions answered by any two students in different classes is also different, so each class must have a unique score.

Suppose we arrange the classes in decreasing order of scores. To maximize the number of students who only answered one question, we should minimize the scores of the other students. Let x be the number of classes with a score of 1. Since each student must answer at least one question, the lowest possible unique scores for the remaining classes would be 2, 3, 4, and 5.

The total number of questions solved is 40. So, we have the equation:

x * 1 + (6 * 2) + (6 * 3) + (6 * 4) + (6 * 5) = 40 + 20x

Now we can solve for x:

20x = 40 - (6 * 2 + 6 * 3 + 6 * 4 + 6 * 5)
20x = 40 - (12 + 18 + 24 + 30)
20x = 40 - 84
20x = -44

Since we cannot have a negative number of classes, this approach doesn't work. Instead, let's try maximizing the number of classes with a score of 2:

y * 2 + (6 * 3) + (6 * 4) + (6 * 5) + (6 * 6) = 40 + 10y

Now we can solve for y:

10y = 40 - (18 + 24 + 30 + 36)
10y = 40 - 108
10y = -68

Again, we cannot have a negative number of classes. We must adjust our approach by increasing the score of the first class while keeping the other scores minimal:

(6 * 1) + (6 * 2) + (6 * 3) + (6 * 4) + z * 5 = 40

Now we can solve for z:

z * 5 = 40 - (6 + 12 + 18 + 24)
z * 5 = 40 - 60
z * 5 = -20
z = -4

Since we cannot have a negative number of students, the initial assumption of having a class with a score of 1 is not valid. The minimum possible scores for each class should be 2, 3, 4, and 5, and the remaining students in the last class will have a score of 1. In this case, the last class will have only one student scoring 1, and the other students in that class will score at least 6.

Final answer: There is a maximum of 1 student who only answered one question.
assistant