- #1

Math100

- 772

- 219

- Homework Statement
- Solve the puzzle-problem below:

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.

Then we have 3x+2y+0.5z=100

such that x+y+z=100.

Note that x+y+z=100 gives us z=100-x-y.

Substituting this result into 3x+2y+0.5z=100 and multiplying it

by 2 produces: 5x+3y=100.

Consider the Diophantine equation 5x+3y=100.

Applying the Euclidean Algorithm produces:

5=1(3)+2

3=1(2)+1

2=2(1)+0.

Now we have 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(5-3)

=2(3)-1(5).

This means 100=100[2(3)-1(5)]

=200(3)-100(5).

Thus, xo=200 and yo=-100.

All solutions in the integers are determined by:

x=200+(3/1)t=200+3t for some integer t,

y=-100-(5/1)t=-100-5t for some integer t.

Therefore, x=200+3t and y=-100-5t.

To find all solutions in the positive integers of the

Diophantine equation 5x+3y=100, we solve for t:

x=200+3t##\geq##0 ##\land## y=-100-5t##\geq##0

t##\geq##-200/3 ##\land## t##\leq##-20.

Now I'm stuck. I know I need to combine these two inequalities and solve for integers t in order to find the solution. But it seems like after I solve the inequality, I found out that the integers t are negative values. Can anyone please find where my mistake(s) are in this problem? Thank you.

and z be the number of children.

Then we have 3x+2y+0.5z=100

such that x+y+z=100.

Note that x+y+z=100 gives us z=100-x-y.

Substituting this result into 3x+2y+0.5z=100 and multiplying it

by 2 produces: 5x+3y=100.

Consider the Diophantine equation 5x+3y=100.

Applying the Euclidean Algorithm produces:

5=1(3)+2

3=1(2)+1

2=2(1)+0.

Now we have 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(5-3)

=2(3)-1(5).

This means 100=100[2(3)-1(5)]

=200(3)-100(5).

Thus, xo=200 and yo=-100.

All solutions in the integers are determined by:

x=200+(3/1)t=200+3t for some integer t,

y=-100-(5/1)t=-100-5t for some integer t.

Therefore, x=200+3t and y=-100-5t.

To find all solutions in the positive integers of the

Diophantine equation 5x+3y=100, we solve for t:

x=200+3t##\geq##0 ##\land## y=-100-5t##\geq##0

t##\geq##-200/3 ##\land## t##\leq##-20.

Now I'm stuck. I know I need to combine these two inequalities and solve for integers t in order to find the solution. But it seems like after I solve the inequality, I found out that the integers t are negative values. Can anyone please find where my mistake(s) are in this problem? Thank you.