¶ 2023 AIME I Problem 14
Problem:
The following analog clock has two hands that can move independently of each other.
Initially, both hands point to the number . The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Let be the number of sequences of hand movements such that during the sequence, every possible positioning of the hands appears exactly once, and at the end of the movements, the hands have returned to their initial position. Find the remainder when is divided by .
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Solution:
The key observation in this problem is to model the problem as the following:
Considering "Grid Walking problems", we have these conditions:
Suppose that the hour hand is the x coordinate, and the minute hand is the y coordinate, and we remember that these hands move completely independently.
Each movement is to the up and right, where things would wrap around.
The key observation in this problem is to consider fixing the total number of movements "right" that are made.
Consider an arbitrary coordinate , We would have been only at from or , taken particularly mod 12 to account for "wrapping around".
Then, we see that from to , we'd need to go "horizontally", and would not be able to touch , and so this helps show how the path can be chosen & created specifically based off the total number of moves right & up, that add up to 12 total steps.
We also notice how choosing numbers that are not relatively prime such as and results in a path that prematurely meets the origin, and so the only valid solutions would be such that these are both relatively prime, and a denotes right moves and b denotes up moves.
Thus, our answer would be the number of solutions of paths , , and , so our answer would be
The problems on this page are the property of the MAA's American Mathematics Competitions