- #1
Math100
- 813
- 229
- 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.
