Difference between revisions of "2022 AMC 8 Problems/Problem 25"
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− | There are always three possible leaves to jump to every time the cricket decides to jump, so there | + | There are always three possible leaves to jump to every time the cricket decides to jump, so there is a total number of <math> 3^4 </math> routes. Let <math>A</math> denote the leaf cricket starts at, and <math>B, C, D</math> be the other leaves. If we want the cricket to move to leaf <math> A </math> for its last jump, the cricket cannot jump to leaf <math> A </math> for its third jump. Also, considering that the cricket starts at leaf <math> A </math>, he cannot jump to leaf <math> A </math> for its first jump. Note that there are <math> 3\cdot2=6 </math> paths if the cricket moves to leaf <math> A </math> for its third jump. Therefore, we can conclude that the total number of possible paths for the cricket to return to leaf <math> A </math> after four jumps is <math> 3^3 - 6 = 21 </math>, so the answer is <math> \frac{21}{3^4} = \frac{21}{81}=\boxed{\textbf{(E) }\frac{7}{27}} </math>. |
~[[User:Bloggish|Bloggish]] | ~[[User:Bloggish|Bloggish]] |
Revision as of 20:19, 28 January 2023
Contents
- 1 Problem
- 2 Solution 1 (Casework)
- 3 Solution 2 (Casework)
- 4 Solution 3 (Complement)
- 5 Solution 4 (Recursion)
- 6 Solution 5 (Dynamic Programming)
- 7 Solution 6 (Generating Function)
- 8 Remark
- 9 Video Solution
- 10 Video Solution by OmegaLearn
- 11 Video Solution
- 12 Video Solution
- 13 Video Solution
- 14 Video Solution
- 15 Video Solution
- 16 See Also
Problem
A cricket randomly hops between leaves, on each turn hopping to one of the other
leaves with equal probability. After
hops what is the probability that the cricket has returned to the leaf where it started?
Solution 1 (Casework)
Let denote the leaf where the cricket starts and
denote one of the other
leaves. Note that:
- If the cricket is at
then the probability that it hops to
next is
- If the cricket is at
then the probability that it hops to
next is
- If the cricket is at
then the probability that it hops to
next is
We apply casework to the possible paths of the cricket:
The probability for this case is
The probability for this case is
Together, the probability that the cricket returns to after
hops is
~MRENTHUSIASM
Solution 2 (Casework)
We can label the leaves as shown below:
Carefully counting cases, we see that there are ways for the cricket to return to leaf
after four hops if its first hop was to leaf
:
By symmetry, we know that there are ways if the cricket's first hop was to leaf
, and there are
ways if the cricket's first hop was to leaf
. So, there are
ways in total for the cricket to return to leaf
after four hops.
Since there are possible ways altogether for the cricket to hop to any other leaf four times, the answer is
.
~mahaler
Solution 3 (Complement)
There are always three possible leaves to jump to every time the cricket decides to jump, so there is a total number of routes. Let
denote the leaf cricket starts at, and
be the other leaves. If we want the cricket to move to leaf
for its last jump, the cricket cannot jump to leaf
for its third jump. Also, considering that the cricket starts at leaf
, he cannot jump to leaf
for its first jump. Note that there are
paths if the cricket moves to leaf
for its third jump. Therefore, we can conclude that the total number of possible paths for the cricket to return to leaf
after four jumps is
, so the answer is
.
Solution 4 (Recursion)
Denote to be the probability that the cricket would return back to the first point after
hops. Then, we get the recursive formula
because if the leaf is not on the target leaf, then there is a
probability that it will make it back.
With this formula and the fact that (After one hop, the cricket can never be back to the target leaf.), we have
so our answer is
.
~wamofan
Solution 5 (Dynamic Programming)
Let denote the leaf cricket starts at, and
be the other leaves, similar to Solution 2.
Let be the probability the cricket lands on
after
hops,
be the probability the cricket lands on
after crawling
hops, and etc.
Note that and
For
the probability that the cricket land on each leaf after
hops is
the sum of the probability the cricket land on other leaves after
hops. So, we have
It follows that
We construct the following table:
Therefore, the answer is
.
Solution 6 (Generating Function)
Assign the leaves to and
modulo
and let
be the starting leaf. We then use generating functions with relation to the change of leaves. For example, from
to
would be a change of
and from
to
would be a change of
This generating function is equal to
It is clear that we want the coefficients in the form of
where
is a positive integer. One application of roots of unity filter gives us a successful case count of
Therefore, the answer is
~sigma
Remark
This problem is a reduced version of 1985 AIME Problem 12, changing steps into
steps.
This problem is also similar to 2003 AIME II Problem 13.
Video Solution
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Video Solution by OmegaLearn
https://youtu.be/kE15Sy0B2Pk?t=633
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=85A6av3oqRo
~Mathematical Dexterity
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=2588
~Interstigation
Video Solution
https://www.youtube.com/watch?v=H1zxrkq6DKg
Video Solution
https://youtu.be/0orAAUaLIO0?t=609
~STEMbreezy
Video Solution
~savannahsolver
See Also
2022 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.