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Actually I put answer to wrong question before. The answer shouldn't be C.

(1) More than one value of x and y that satisfy this condition.
(2) Way way many values of x and y that satisfy this condition.
(combined) x = 25, y = 16 (just another subset of what Stmnt2 gives us, so Stmnt2 is sufficient)

Last edited by wonder_gmat on 22 Dec 2003, 09:59, edited 1 time in total.

In MO when x=4,z=2,y=1 not -1 since after transformation of b) (z-1)^2=y and when z=2 then y=1. 1 raised to any power is 1 so 1 is always perfect square. Can we assume that in this case B) is correct?

In MO when x=4,z=2,y=1 not -1 since after transformation of b) (z-1)^2=y and when z=2 then y=1. 1 raised to any power is 1 so 1 is always perfect square. Can we assume that in this case B) is correct?

yes, (sqrt(X)-1)^2 or (Z-1)^2 = Y, will hold good, always.

1) rearrange as z^2+y=8z+1 =z^2-8z+y-1=0 = z(z-8)+y-1=0

choose a few numbers starting with x x=100; therefore z=10 thus: 10(2)+21-1=0 y = 21 and x=100 (no) x=25 z=5 5(-3)+16-1=0 y=16 x=25 (yes)

Insufficient

2) x-y=2z-1

rewrite: z^2-2z-y+1=0

z(z-2)-y+1=0 x=121 z=11; thus y=100 (yes consecutive) x= 100 z =10 y=81 (yes consecutive) x=16 z=4 y= 9 (yes consecutive) x=81 z=9 y=64 (yes x and y are consecutive square numbers) x=4 z=2 y=-1 (doesn't apply b/c x,y,z are all positive integers)

Sufficient

B

titliest,

could you check the bold part. the solution does not equal 0.

x= 25 , y =16 is the only solution.

OOPS! Yes - you're right Praetorian = X=25, Y=16 is the only solution - I agree with DJ - it is tough doing this in the office.

I stand corrected again! The answer is still B - I was right by being wrong! As DJ pointed out, X=36, Z=6, Y=13 (This is a solution but X and Y are not consecutive squares).

Last edited by Titleist on 22 Dec 2003, 12:10, edited 2 times in total.

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