2006 AMC 12B Problems/Problem 18

Problem

An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?

$\mathrm{(A)}\ 120 \qquad \mathrm{(B)}\ 121 \qquad \mathrm{(C)}\ 221 \qquad \mathrm{(D)}\ 230 \qquad \mathrm{(E)}\ 231$

Solution

Let the starting point be $(0,0)$ . After $10$ steps we can only be in locations $(x,y)$ where $|x|+|y|\leq 10$ . Additionally, each step changes the parity of exactly one coordinate. Hence after $10$ steps we can only be in locations $(x,y)$ where $x+y$ is even. It can easily be shown that each location that satisfies these two conditions is indeed reachable.

Once we pick $x\in\{-10,\dots,10\}$ , we have $11-|x|$ valid choices for $y$ , giving a total of $\boxed{121}$ possible positions.

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

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2006 AMC 12B Problems/Problem 18

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