Difference between revisions of "2024 AMC 8 Problems/Problem 23"
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+ | Let <math>f(x, y)</math> be the number of cells the line segment from <math>(0, 0)</math> to <math>(x, y)</math> passes through. The problem is equivalent to finding <math>f(5000-2000, 8000-3000)=f(3000, 5000)</math> times. Sometimes the segment passes through lattice points in between the endpoints, which happens <math>\text{gcd}(3000, 5000)-1=999</math> times. This partitions the segment into <math>1000</math> congruent pieces that pass through <math>f(3, 5)</math> cells, which means the answer is <math>1000f(3, 5).</math> Note that a new square is entered when the lines pass through one of the lines in the coordinate grid, which for <math>f(3, 5)</math> happens <math>3-1+5-1=6</math> times. Because <math>3</math> and <math>5</math> are relatively prime, no lattice point except for the endpoints intersects the line segment from <math>(0, 0)</math> to <math>(3, 5).</math> This means that including the first cell closest to <math>(0, 0),</math> The segment passes through <math>f(3, 5)=6+1=7</math> cells. Thus, the answer is <math>7000.</math> |
Revision as of 16:51, 25 January 2024
Problem
Solution 1
Let be the number of cells the line segment from
to
passes through. The problem is equivalent to finding
times. Sometimes the segment passes through lattice points in between the endpoints, which happens
times. This partitions the segment into
congruent pieces that pass through
cells, which means the answer is
Note that a new square is entered when the lines pass through one of the lines in the coordinate grid, which for
happens
times. Because
and
are relatively prime, no lattice point except for the endpoints intersects the line segment from
to
This means that including the first cell closest to
The segment passes through
cells. Thus, the answer is