2024 AMC 8 Problems/Problem 23

Revision as of 15:51, 25 January 2024 by Bs2012 (talk | contribs) (Solution 1)

Problem

Solution 1

Let $f(x, y)$ be the number of cells the line segment from $(0, 0)$ to $(x, y)$ passes through. The problem is equivalent to finding $f(5000-2000, 8000-3000)=f(3000, 5000)$ times. 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 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 $7000.$