Difference between revisions of "2023 AIME I Problems/Problem 14"
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The following analog clock has two hands that can move independently of each other. | The following analog clock has two hands that can move independently of each other. | ||
<asy> | <asy> | ||
− | unitsize(2cm); | + | unitsize(2cm); |
− | draw(unitcircle,black+linewidth(2)); | + | draw(unitcircle,black+linewidth(2)); |
− | for (int i = 0; i < 12; ++i) { | + | |
− | draw(0.9*dir(30*i)--dir(30*i)); | + | for (int i = 0; i < 12; ++i) { |
− | } | + | draw(0.9*dir(30*i)--dir(30*i)); |
− | for (int i = 0; i < 4; ++i) { | + | } |
− | draw(0.85*dir(90*i)--dir(90*i),black+linewidth(2)); | + | for (int i = 0; i < 4; ++i) { |
− | } | + | draw(0.85*dir(90*i)--dir(90*i),black+linewidth(2)); |
− | for (int i = | + | } |
− | label("\small" + (string) i, dir(90 - i * 30) * 0.75); | + | for (int i = 1; i < 13; ++i) { |
− | } | + | label("\small" + (string) i, dir(90 - i * 30) * 0.75); |
− | draw((0,0)--0.6*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); | + | } |
− | draw((0,0)--0.4*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); | + | draw((0,0)--0.6*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); |
+ | draw((0,0)--0.4*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); | ||
</asy> | </asy> | ||
Initially, both hands point to the number <math>12</math>. 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. | Initially, both hands point to the number <math>12</math>. 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. | ||
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==Solution 1 (Drawing the Grid)== | ==Solution 1 (Drawing the Grid)== | ||
− | This problem is, in essence, the following: A <math>12\times12</math> 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 <math>(0,0)</math> in the coordinate plane. Every second, he jumps either to the right or up, until he reaches an <math>x</math>- or <math>y</math>-coordinate of <math>12</math>. At this point, if he tries to jump to a coordinate outside the square from <math>(0,0)</math> to <math>(11,11)</math>, he "wraps around" and ends up at an <math>x</math>- or <math>y</math>- coordinate of <math>0</math>. How many ways are there for Felix to jump on every grid point in this square, so that he ends at <math>(0,0)</math>? This is consistent with the construction of the flat torus as <math>\mathbb Z^2/12\mathbb Z^2</math> (2-dimensional modular arithmetic. <math>(\mathbb{Z}_{12})^2</math> | + | This problem is, in essence, the following: A <math>12\times12</math> 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 <math>(0,0)</math> in the coordinate plane. Every second, he jumps either to the right or up, until he reaches an <math>x</math>- or <math>y</math>-coordinate of <math>12</math>. At this point, if he tries to jump to a coordinate outside the square from <math>(0,0)</math> to <math>(11,11)</math>, he "wraps around" and ends up at an <math>x</math>- or <math>y</math>- coordinate of <math>0</math>. How many ways are there for Felix to jump on every grid point in this square, so that he ends at <math>(0,0)</math>? This is consistent with the construction of the flat torus as <math>\mathbb Z^2/12\mathbb Z^2</math> (2-dimensional modular arithmetic. <math>(\mathbb{Z}_{12})^2</math>) |
Moving on, define a <math>\textit{path}</math> from point <math>A</math> to point <math>B</math> to be a sequence of "up"s and "right"s that takes Felix from <math>A</math> to <math>B</math>. The <math>\textit{distance}</math> from <math>A</math> to <math>B</math> is the length of the shortest path from <math>A</math> to <math>B</math>. At the crux of this problem is the following consideration: The points <math>A_i=(i,12-i), i\in{0,...,11}</math> are pairwise equidistant, each pair having distance of <math>12</math> in both directions. | Moving on, define a <math>\textit{path}</math> from point <math>A</math> to point <math>B</math> to be a sequence of "up"s and "right"s that takes Felix from <math>A</math> to <math>B</math>. The <math>\textit{distance}</math> from <math>A</math> to <math>B</math> is the length of the shortest path from <math>A</math> to <math>B</math>. At the crux of this problem is the following consideration: The points <math>A_i=(i,12-i), i\in{0,...,11}</math> are pairwise equidistant, each pair having distance of <math>12</math> in both directions. | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/tsoLZj8Dz2o | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
Latest revision as of 22:01, 4 January 2024
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
~MathProblemSolvingSkills.com
Animated Video Solution
~Star League (https://starleague.us)
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.