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Franklin the fly starts at the point $(0,0)$ in the coordinate plane. At each point, Franklin takes a step to the right, left, up, or down. After $12$ steps, how many different points could Franklin end up at?

User Roheen
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Franklin's path, reduced to even moves, allows endpoints (x, y) where x + y is even, and |x| + |y| is no greater than 10. Calculations show 9 possibilities for |x| + |y| ≤ 2 and 121 possibilities for |x| + |y| ≤ 10, forming respective 3x3 and 11x11 grids.

Consider Franklin's random 10-step path, denoted as NESWNNESEW (N for North, S for South, W for West, and E for East). To find the ending point, we can reduce the path by canceling each E-W and N-S pair without changing the endpoint. In this example, the path reduces to NE. It is crucial to note that the reduced path must have an even number of moves. Thus, any endpoint (x, y) is possible where x + y is even, and |x| + |y| is no greater than 10.

Now, examine the possibilities for |x| + |y|:

If |x| + |y| = 0, there is only one possibility for (x, y) at the origin (0, 0).

If |x| + |y| ≤ 2, there are 9 possible paths: (0,0), (-1,-1), (-1,0), (-1,1), (0,-1), (0,1), (1,-1), (1,0), (1,1). These solutions form a 3x3 diagonal grid.

The solutions with |x| + |y| ≤ 10 form an 11x11 grid, resulting in 11^2 = 121 possible ways.

In summary, the endpoint possibilities are determined by the reduced path, leading to specific conditions for |x| + |y|. Calculations reveal a 3x3 grid for |x| + |y| ≤ 2 and an 11x11 grid for |x| + |y| ≤ 10, yielding a total of 121 possible ways.

User Neal Kruis
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