Giải:

Call the number of students who have won all `3` subjects a (student) 

Call the number of students who have won both subjects b (students) 

Call the number of students who have won only `1` subject c (student) 

The total number of prizes achieved is: `3 x a + 2 x b + c = 15` (prizes). 

Because the total number of students won `3` prizes, `2` prizes, `1` prize gradually increased, `a < b < c`. Since any of the two subjects have at least one student who has won both Literature and Foreign Languages, so: - At least `1` student has won both Literature and Mathematics. - There is at least `1` student who has won both Literature and Foreign Languages. - There is at least 1 student who has won both Literature and Foreign Languages. So `b = 3`. Suppose `a = 2`, the youngest `b` is 3, the smallest is 4; therefore the total number of the smallest prizes is: `3 x 2 + 2 x 3 + 4 = 16 > 15` (type). Therefore `a < 2`, so `a = 1`.

We have: `3 x 1 + 2 x b + c = 15` infer: `2 x b + c = 12`. 

If `b = 3` then `c = 12 - 2 x 3 = 6` (correct). 

If `b = 4` then `c = 12 - 2 x 4 = 4` (type because it is contrary to conditions b < c) 

So there is `1` friend who won `3` prizes, `3` you won `2` prizes, `6` you won `1` prize. 

That team has a student count of: `1 + 3 + 6 = 10` (you).