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 15:51, 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