Question 53523
<pre><font size = 6><b>Mr. Smith has $100.00 to buy 100 
animals with. Cows cost $10.00 each, 
pigs cost $3.00 each and chickens 
cost $.50 each. How many can he buy 
with exactly $100.00.  I have this 
equation so far, but where do I go 
from here? 

10x + 3y + .5z = 100

Multiply through by 10 to clear decimal:

100x + 30y + 5z = 1000

Divide by 5

20x + 6y + z = 200
   x + y + z = 100

19x + 5y = 100

19x = 100 - 5y

19x = 5(20 - y)

  x = 5(20 - y)/19

Since x is a positive integer,
the 19 must divide evenly into
5(20 - y).  the 19 will not
divide evenly into the 5, so
19 must divide into 20 - y.

20 - y is at most 19, since
y must be at least 1.  The
only integer at least 19
which is divisible by 19 is
19 itself.  Therefore

20 - y = 19
    -y = -1
     y = 1 
 
x = 5(20 - y)/19
x = 5(20 - 1)/19
x = 5(19)/19
x = 5

x + y + z = 100
5 + 1 + z = 100
    6 + z = 100
        z = 94

Therefore he bought
5 cows, 1 pig, 
and 94 chickens.

Edwin</pre>