Difference between revisions of "2011 AMC 12B Problems/Problem 11"
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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>. | ||
− | Thus, the answer is <math> = \{3 \textbf{ | + | Thus, the answer is <math> = \boxed{3 \textbf{}} </math>. |
== See also == | == See also == |
Latest revision as of 10:42, 24 February 2021
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 Debapriya 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 (based off the 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 is impossible for him to go from to in an even number of moves. Therefore, Debapriya 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 |
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