AMC8 2024

Let $f(x, y)$ be the number of cells the line segment from $(0, 0)$ to $(x, y)$ passes through. The problem is then equivalent to finding \[f(5000-2000, 8000-3000)=f(3000, 5000).\] Sometimes the segment passes through lattice points in between the endpoints, which happens $\text{gcd}(3000, 5000)-1=999$ times. This partitions the segment into $1000$ congruent pieces that each pass through $f(3, 5)$ cells, which means the answer is \[1000f(3, 5).\] Note that a new square is entered when the lines pass through one of the lines in the coordinate grid, which for $f(3, 5)$ happens $3-1+5-1=6$ times. Because $3$ and $5$ are relatively prime, no lattice point except for the endpoints intersects the line segment from $(0, 0)$ to $(3, 5).$ This means that including the first cell closest to $(0, 0),$ The segment passes through $f(3, 5)=6+1=7$ cells. Thus, the answer is $\boxed{\textbf{(C)}7000}.$ Alternatively, $f(3, 5)$ can be found by drawing an accurate diagram, leaving you with the same answer. Note: A general form for finding $f(x, y)$ is $x+y-\text{gcd}(x, y).$ We subtract $\text{gcd}(x, y)$ to account for overlapping, when the line segment goes through a lattice point.