Is it possible to solve this, make a formula/algorithm?

Question

Algorithms

Problem about monks
There are 3 groups of monks: leading, simple, students. M monks eat N pies by order.
Leader in N1 pies, simple in N2 pies, students in N3 pies (N, N1, N2, N3 - can be whole and fractional numbers).
How many monks were: leading (M1), simple (M2) and disciples (M3).
Input data
M - number of monks in total
N - number of pies in total
N1 - number of pies eaten by one leading monk
N2 - number of pies eaten by one simple monk
N2 - number of pies eaten by one student monk
Results
M1 - number of leading monks
M2 - number of simple monks
M3 - the number of monks-apprentices

Answer 1

Для каждого M1 от 0 до N/N1
    M2 = (N - M1*N1 -M*N3 + M1*N3) / (N2 - N3)
    M3 = M - M1 - M2
        если M2 целое число и М3>0
            то вернуть ответ М1, М2, М3

Answer 2

In this form, it is impossible to solve, since the condition does not say how the output data depends on the input. And it is not specified what the variables M and P mean.

Answer 3

It is possible with two nested loops - we sort through M1 and M2, and M3 is found from the system of equations.
M1 + M2 + M3 = M
M1*P1 + M2*P2 + M3*P3 = P

Answer 4

here the question is in the condition, if you read literally, then the system of equations is as follows:
N = N1*M1 + N2*M2 +N3*M3
M = M1 + M2 + M3
M1>=0
M2>=0
M3>=0
obviously - it can be solve, but there can be more than 1 solution
trivial algorithm - enumeration of all possible M1,M2,M3