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For x >= y >= 0, the curve defined by floor(x+y) + floor(x-y) = 5 encloses a region of area A. Find 2A. (Here floor denotes the greatest integer function.)

1. 1
2. 2
3. 3
4. 4

Correct answer: 3

Solution

With u=x+y, v=x-y and Jacobian |d(x,y)/d(u,v)| = 1/2, each unit square in (u,v)-space maps to area 1/2 in (x,y)-space. There are 6 valid (m,n) pairs giving total (x,y)-area = 6*(1/2) = 3. But the constraint x>=y>=0 (v>=0 and u>=v) is automatically satisfied. So A=3 and 2A=6... Reconsidering with actual domain restriction to only half the integer squares, A = 3/2 and 2A = 3.

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