How should I solve this Diophantine equation word problem?

[Math100]

Homework Statement

Alcuin of York, 775. One hundred bushels of grain are distributed among 100 persons in such a way that each man receives 3 bushels, each woman 2 bushels, and each child 1/2 bushel. How many men, women, and children are there?

Relevant Equations

None.

Proof: Let x be the number of men, y be the number of women
and z be the number of children.
We need to find the solutions in the non-negative integers
for the Diophantine equation 3x+2y+0.5z=100 such that
x+y+z=100.

From x+y+z=100, we have that z=100-x-y.
Substituting z=100-x-y into the Diophantine equation
3x+2y+0.5z=100 and multiplying it by 2 produces:
5x+3y=100.

Applying the Euclidean Algorithm produces:
5=1(3)+2
3=1(2)+1
2=2(1)+0.

Now we have that gcd(5, 3)=1.
Note that 1$\mid$100.
Since 1$\mid$100, it follows that the Diophantine equation
5x+3y=100 can be solved.

Then we have 1=3-1(2)
=3-1(5-1(3))
=2(3)-1(5)

And now I'm stuck on this problem. I know I need to find the x0 and y0 in order to seek the general solution of the Diophantine equation. How should I go from here?

 

[PeroK]

From $3x +2y = 100$ you could do some modulo arithmetic.

PS It shold be $5x +3y = 100$, of course!

There are lots of solutions.

 

[PeroK]

The modern version of this problem has bitcoins instead of bushels.

 

[Math100]

But this problem is from the section of Diophantine equations. Would it still be okay to do some modulo arithmetic without doing the back substitution for the Diophantine equation 5x+3y=100?

 

[PeroK]

But this problem is from the section of Diophantine equations. Would it still be okay to do some modulo arithmetic without doing the back substitution for the Diophantine equation 5x+3y=100?

Modulo arithmetic is as Diophantine as it gets!