Difference between revisions of "2011 AMC 12B Problems/Problem 11"
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− | Since the frog always jumps in length <math>5</math> and lands on a lattice point, the sum of its coordinates must change either by <math>5</math> (by jumping parallel to the x- or y-axis), or by <math>3</math> or <math>4</math> ( | + | Since the frog always jumps in length <math>5</math> and lands on a lattice point, the sum of its coordinates must change either by <math>5</math> (by jumping parallel to the x- or y-axis), or by <math>3</math> or <math>4</math> (3-4-5 right triangle). |
− | Because either <math>1</math>, <math>5</math>, or <math>7</math> is always the change of the sum of the coordinates, the sum of the coordinates will always change from odd to even or vice versa. | + | Because either <math>1</math>, <math>5</math>, or <math>7</math> is always the change of the sum of the coordinates, the sum of the coordinates will always change from odd to even or vice versa. Thus, it can't go from <math>(0,0)</math> to <math>(1,0)</math> in an even number of moves. Therefore, the frog cannot reach <math>(1,0)</math> in two moves. |
However, a path is possible in 3 moves: from <math>(0,0)</math> to <math>(3,4)</math> to <math>(6,0)</math> to <math>(1,0)</math>. | However, a path is possible in 3 moves: from <math>(0,0)</math> to <math>(3,4)</math> to <math>(6,0)</math> to <math>(1,0)</math>. |
Latest revision as of 18:39, 7 August 2023
Problem
A frog located at , with both and integers, makes successive jumps of length and always lands on points with integer coordinates. Suppose that the frog starts at and ends at . What is the smallest possible number of jumps the frog makes?
Solution
Since the frog always jumps in length and lands on a lattice point, the sum of its coordinates must change either by (by jumping parallel to the x- or y-axis), or by or (3-4-5 right triangle).
Because either , , or is always the change of the sum of the coordinates, the sum of the coordinates will always change from odd to even or vice versa. Thus, it can't go from to in an even number of moves. Therefore, the frog cannot reach in two moves.
However, a path is possible in 3 moves: from to to to .
Thus, the answer is .
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.