Difference between revisions of "User:John0512"
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− | + | ==2021 AMC 8== | |
+ | 2021 AMC 8 problems and solutions. The test has not been held, and will never be held. | ||
+ | |||
+ | ==Problems== | ||
+ | ERROR: Content not found | ||
+ | ==Solutions== | ||
+ | ERROR: Content not found | ||
+ | ==Results== | ||
+ | |||
+ | Highest Score: 0.00 | ||
+ | |||
+ | Distinguished Honor Roll: 0.00 | ||
+ | |||
+ | Honor Roll: 0.00 | ||
+ | |||
+ | Average Score: 0.00 | ||
+ | |||
+ | Standard Deviation: 0.00 | ||
+ | |||
+ | ==Unnamed Theorem== | ||
+ | |||
+ | I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before). | ||
+ | |||
+ | Claim: Given a set <math>S=\{1,2,3\cdots n\}</math> where <math>n</math> is a positive integer, the number of ways to choose a subset of <math>S</math> then permute said subset is <math>\lfloor n!\cdot e\rfloor.</math> | ||
+ | |||
+ | Proof: The number of ways to choose a subset of size <math>i</math> and then permute it is <math>\binom{n}{i}\cdot i!</math>. Therefore, the number of ways to choose any subset of <math>S</math> is <cmath>\sum_{i=0}^n \binom{n}{i}\cdot i! = \sum_{i=0}^n \frac{n!}{(n-i)!}.</cmath> This is also equal to <math>\sum_{i=0}^n \frac{n!}{i!}</math> by symmetry across <math>i=\frac{n}{2}</math>. This is also <math>n! \cdot \sum_{i=0}^n \frac{1}{i!}.</math> Note that <math>e</math> is defined as <math>\sum_{i=0}^\infty \frac{1}{i!}</math>, so our expression becomes <cmath>n!(e-\sum_{i={n+1}}^\infty \frac{1}{i!}).</cmath> We claim that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{n!}</math> for all positive integers <math>n</math>. | ||
+ | |||
+ | Since the reciprocal of a factorial decreases faster than a geometric series, we have that <math>\sum_{i={n+1}}^\infty \frac{1}{i!}<\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+1)}+\frac{1}{(n+1)!(n+1)^2}\cdots</math>. The right side we can evaluate as <math>\frac{1}{n(n!)}</math>, which is always less than or equal to <math>\frac{1}{n!}</math>. This means that the terms being subtracted are always strictly less than <math>\frac{1}{n!}</math>, so we can simply write it as <cmath>\lfloor n!\cdot e\rfloor.</cmath> | ||
+ | |||
+ | Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order? | ||
+ | |||
+ | Solution to example: This is equivalent to the Unnamed Theorem for <math>n=5</math>, so our answer is <math>\lfloor 120e \rfloor=\boxed{326}</math>. | ||
+ | |||
+ | Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is <math>\boxed{1}</math>. Note that this contradicts with the answer given by the Unnamed Theorem. |
Latest revision as of 03:03, 12 March 2023
2021 AMC 8
2021 AMC 8 problems and solutions. The test has not been held, and will never be held.
Problems
ERROR: Content not found
Solutions
ERROR: Content not found
Results
Highest Score: 0.00
Distinguished Honor Roll: 0.00
Honor Roll: 0.00
Average Score: 0.00
Standard Deviation: 0.00
Unnamed Theorem
I have something called the Unnamed Theorem (which I did not name as I have not confirmed that this theorem has not existed before).
Claim: Given a set where
is a positive integer, the number of ways to choose a subset of
then permute said subset is
Proof: The number of ways to choose a subset of size and then permute it is
. Therefore, the number of ways to choose any subset of
is
This is also equal to
by symmetry across
. This is also
Note that
is defined as
, so our expression becomes
We claim that
for all positive integers
.
Since the reciprocal of a factorial decreases faster than a geometric series, we have that . The right side we can evaluate as
, which is always less than or equal to
. This means that the terms being subtracted are always strictly less than
, so we can simply write it as
Example: How many ways are there 5 distinct clones of mathicorn to each either accept or reject me, then for me to go through the ones that accepted me in some order?
Solution to example: This is equivalent to the Unnamed Theorem for , so our answer is
.
Solution 2: Since I am not orz, all 5 clones will reject me, so the answer is . Note that this contradicts with the answer given by the Unnamed Theorem.