Difference between revisions of "2011 AMC 12B Problems/Problem 11"
(→Solution) |
(→Solution) |
||
| Line 11: | Line 11: | ||
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> = \boxed{3 \textbf{( | + | Thus, the answer is <math> = \boxed{3 \textbf{()}} </math>. |
== See also == | == See also == | ||
Revision as of 09:42, 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
