AMC12 2006 B

AMC12 2006 B · Q18

AMC12 2006 B · Q18. It mainly tests Basic counting (rules of product/sum), Casework.

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?

平面上的一个物体从一个格点移动到另一个格点。每一步，物体可以向右移动一单位、向左移动一单位、向上移动一单位或向下移动一单位。如果物体从原点开始，走一条十步路径，可能的终点有多少个不同的点？

(A) 120 120

(B) 121 121

(D) 230 230

(E) 231 231

Answer

Correct choice: (B)

正确答案：（B）

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.

设起点为 $(0,0)$。走 $10$ 步后，只可能到达满足 $|x|+|y|\leq 10$ 的位置 $(x,y)$。此外，每一步恰好改变一个坐标的奇偶性。因此走 $10$ 步后，只可能到达满足 $x+y$ 为偶数的位置 $(x,y)$。容易证明，满足这两个条件的每个位置确实都可达。一旦选定 $x\in\{-10,\dots,10\}$，则 $y$ 有 $11-|x|$ 个有效选择，因此总共有 $\boxed{121}$ 个可能位置。

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