Difference between revisions of "2011 AMC 12B Problems/Problem 11"
(→Solution: Fixed solution) |
Megaboy6679 (talk | contribs) m |
||
| (12 intermediate revisions by 9 users not shown) | |||
| Line 1: | Line 1: | ||
| − | ==Problem== | + | == Problem == |
A frog located at <math>(x,y)</math>, with both <math>x</math> and <math>y</math> integers, makes successive jumps of length <math>5</math> and always lands on points with integer coordinates. Suppose that the frog starts at <math>(0,0)</math> and ends at <math>(1,0)</math>. What is the smallest possible number of jumps the frog makes? | A frog located at <math>(x,y)</math>, with both <math>x</math> and <math>y</math> integers, makes successive jumps of length <math>5</math> and always lands on points with integer coordinates. Suppose that the frog starts at <math>(0,0)</math> and ends at <math>(1,0)</math>. What is the smallest possible number of jumps the frog makes? | ||
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math> | <math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math> | ||
| + | == Solution == | ||
| + | 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. 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. | |
| − | |||
| − | |||
| − | |||
| − | Because either <math>1</math>, <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>. | 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{}} </math>. | ||
| − | + | == See also == | |
| + | {{AMC12 box|year=2011|ab=B|num-b=10|num-a=12}} | ||
| − | {{ | + | [[Category:Introductory Geometry Problems]] |
| + | {{MAA Notice}} | ||
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. 
