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Difference between revisions of "2011 AMC 12B Problems/Problem 11"

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== Solution ==
 
== Solution ==
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Since Debapriya 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> (based off the 3-4-5 right triangle).
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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> (based off the 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 is impossible for him to go from <math>(0,0)</math> to <math>(1,0)</math> in an even number of moves.  Therefore, Debapriya cannot reach <math>(1,0)</math> in two moves.   
 
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 is impossible for him to go from <math>(0,0)</math> to <math>(1,0)</math> in an even number of moves.  Therefore, Debapriya cannot reach <math>(1,0)</math> in two moves.   

Revision as of 15:47, 12 November 2022

Problem

A frog located at $(x,y)$, with both $x$ and $y$ integers, makes successive jumps of length $5$ and always lands on points with integer coordinates. Suppose that the frog starts at $(0,0)$ and ends at $(1,0)$. What is the smallest possible number of jumps the frog makes?

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

Solution

Since the frog always jumps in length $5$ and lands on a lattice point, the sum of its coordinates must change either by $5$ (by jumping parallel to the x- or y-axis), or by $3$ or $4$ (based off the 3-4-5 right triangle).

Because either $1$, $5$, or $7$ 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 $(0,0)$ to $(1,0)$ in an even number of moves. Therefore, Debapriya cannot reach $(1,0)$ in two moves.

However, a path is possible in 3 moves: from $(0,0)$ to $(3,4)$ to $(6,0)$ to $(1,0)$.

Thus, the answer is $= \boxed{3 \textbf{}}$.

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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