2023 AIME I Problems/Problem 14
Contents
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 .
Solution 1 (Drawing the Grid)
This problem is, in essence, the following: A coordinate grid is placed on a (flat) torus; how many loops are there that pass through each point while only moving up or right? In other words, Felix the frog starts his journey at in the coordinate plane. Every second, he jumps either to the right or up, until he reaches an - or -coordinate of . At this point, if he tries to jump to a coordinate outside the square from to , he "wraps around" and ends up at an - or - coordinate of . How many ways are there for Felix to jump on every grid point in this square, so that he ends at ? This is consistent with the construction of the flat torus as (2-dimensional modular arithmetic.
Moving on, define a from point to point to be a sequence of "up"s and "right"s that takes Felix from to . The from to is the length of the shortest path from to . At the crux of this problem is the following consideration: The points are pairwise equidistant, each pair having distance of in both directions.
A valid complete path then joins two 's, say and . In fact, a link between some and fully determines the rest of the cycle, as the path from must "hug" the path from , to ensure that there are no gaps. We therefore see that if leads to , then leads to . Only the values of relatively prime to result in solutions, though, because otherwise would only lead to . The number of paths from to is , and so the answer is
Notes:
- One can prove that the path from must "hug" the path from by using techniques similar to those in Solution 2.
- One can count the paths as follows: To get from to , Felix takes rights and ups, which can be done in ways.
Solution 2 (Simple Cyclic Permutation Analysis)
This is more of a solution sketch and lacks rigorous proof for interim steps, but illustrates some key observations that lead to a simple solution.
Note that one can visualize this problem as walking on a grid where the edges warp. Your goal is to have a single path across all nodes on the grid leading back to . For convenience, any grid position are presumed to be in .
Note that there are exactly two ways to reach node , namely and .
As a result, if a path includes a step from to , there cannot be a step from to . However, a valid solution must reach , and the only valid step is from .
So a solution that includes a step from to dictates a step from to and by extension steps from to . We observe the equivalent result for steps in the orthogonal direction.
This means that in constructing a valid solution, taking one step in fact dictates N steps, thus it's sufficient to count valid solutions with moves of going right times and times up the grid. The number of distinct solutions can be computed by permuting 2 kinds of indistinguishable objects .
Here we observe, without proof, that if , then we will return to the origin prematurely. For , we only want to count the number of solutions associated with .
(For those attempting a rigorous proof, note that ).
The total number of solutions, noting symmetry, is thus
This yields as our desired answer.
~ cocoa @ https://www.corgillogical.com/
Solution 3 (Matrix Analysis and Permutation)
Define a matrix . Each entry denotes the number of movements the longer hand moves, given that two hands jointly make movements. Thus, the number of movements the shorter hand moves is .
Denote by the remainder of divided by 12. Denote by this remainder matrix.
If two hands can return to their initial positions after 144 movements, then or 11. Denote by (resp. ) the collection of feasible sequences of movements, such that (resp. ).
Define a function , such that for any , the functional value of the entry indexed as is . Thus, function is bijective. This implies .
In the rest of analysis, we count .
We make the following observations:
\begin{enumerate} \item and .
These follow from the definition of .
\item Each column of is a permutation of .
The reasoning is as follows. Suppose there exist , , such that . Then this entails that the positions of two hands after the th movement coincide with their positions after the th movement.
\item For any , is equal to either 0 for all or 1 for all .
The reasoning is as follows. If this does not hold and the th column in is a permutation of , then the th column is no longer a permutation of . This leads to the infeasibility of the movements.
\item for any .
This follows from the conditions that the th column in excluding and the first column in excluding are both permutations of .
\end{enumerate}
All observations jointly imply that . Thus, is a permutation of . Thus, is relatively prime to 12.
Because and , we have , 5, 7, or 11.
Recall that when we move from to , there are 11 steps of movements. Each movement has or 1. Thus, for each given , the number of feasible movements from to is .
Therefore, the total number of feasible movement sequences in this problem is
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
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See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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