Question 143309

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.
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x=18, y=15
x=17, y=16

Any of the above combinations of piles work.
----interesting problem-----
Hope this helps---ptaylor