Question 143309: A man divides his gold coins into two piles. 33 times the difference of these piles is equal to the difference between the squares of each pile. How many in each pile?
Found 2 solutions by ptaylor, ankor@dixie-net.com:
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
Let x=number of pieces in the large pile
And let y=number of pieces in the small pile
33(x-y)=33 times the difference of these piles
x^2-y^2=the difference between the squares of each pile
So our equation to solve is:
33(x-y)=x^2-y^2
Note that x^2-y^2=(x-y)(x+y) so we now have
33(x-y)=(x-y)(x+y) divide each side by (x-y); recognizing that (x-y) not equal to zero
x+y=33
x=32, y=1
x=31, y=2
x=30, y=3
x=29, y=4
etc.
etc.
etc.
---
---
x=18, y=15
x=17, y=16
Any of the above combinations of piles work.
----interesting problem-----
Hope this helps---ptaylor
Answer by ankor@dixie-net.com(22740)  (Show Source):
You can put this solution on YOUR website! A man divides his gold coins into two piles. 33 times the difference of these piles is equal to the difference between the squares of each pile. How many in each pile?
:
Let x = amt on 1 pile
Let y = amt in another pile
:
Write an equation for the statement:
"33 times the difference of these piles is equal to the difference between the squares of each pile."
:
33(x - y) = x^2 - y^2
:
Factor the right side as the difference of squares:
33(x-y) = (x+y)(x-y)
:
divide both sides by (x-y, and we are left with:
33 = x + y
:
Any values of x and y that will satisfy this equation will work:
Three examples
:
x=17,y=16: 33(1) = 17^2 - 16^2 = 33
:
x=20,y=13: 33(7) = 20^2 - 13^2 = 231
:
x=32,y=1: 33(31) = 32^2 - 1^2 = 1023
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