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Difference between revisions of "2024 USAMO Problems/Problem 1"
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Note that if we have a prime number <math>p>n</math> and an integer <math>k>p</math> such that both <math>k</math> and <math>k+1</math> are factors of <math>n!</math>, then the condition cannot be satisfied.
 
Note that if we have a prime number <math>p>n</math> and an integer <math>k>p</math> such that both <math>k</math> and <math>k+1</math> are factors of <math>n!</math>, then the condition cannot be satisfied.
  
−
If <math>n\geq7</math> is odd, then <math>(2)(\frac{n-1}{2})(n-1)=n^2-2n+1</math> is a factor of <math>n!</math>. Also, <math>(n-2)(n)=n^2-2n</math> is a factor of <math>n!</math>. Since <math>2n<n^2-2n</math> for all <math>n\geq7</math>, we can use Bertrand's Postulate to show that there is at least one prime number <math>p</math> such that <math>n<p<n^2-2n</math>. Since we have two consecutive factors of <math>n!</math> and a prime number between the smaller of these factors and <math>n</math>, the condition will not be satisfied for all odd <math>n</math> greater than or equal to <math>7</math>.
+
If <math>n\geq7</math> is odd, then <math>(2)(\frac{n-1}{2})(n-1)=n^2-2n+1</math> is a factor of <math>n!</math>. Also, <math>(n-2)(n)=n^2-2n</math> is a factor of <math>n!</math>. Since <math>2n<n^2-2n</math> for all <math>n\geq7</math>, we can use [[Bertrand's Postulate]] to show that there is at least one prime number <math>p</math> such that <math>n<p<n^2-2n</math>. Since we have two consecutive factors of <math>n!</math> and a prime number between the smaller of these factors and <math>n</math>, the condition will not be satisfied for all odd <math>n</math> greater than or equal to <math>7</math>.
  
 
If <math>n\geq8</math> is even, then <math>(2)(\frac{n-2}{2})(n-2)=n^2-4n+4</math> is a factor of <math>n!</math>. Also, <math>(n-3)(n-1)=n^2-4n+3</math> is a factor of <math>n!</math>. Since <math>2n<n^2-4n+3</math> for all <math>n\geq8</math>, we can use Bertrand's Postulate again to show that there is at least one prime number <math>p</math> such that <math>n<p<n^2-4n+3</math>. Since we have two consecutive factors of <math>n!</math> and a prime number between the smaller of these factors and <math>n</math>, the condition will not be satisfied for all even <math>n</math> greater than or equal to <math>8</math>.
 
If <math>n\geq8</math> is even, then <math>(2)(\frac{n-2}{2})(n-2)=n^2-4n+4</math> is a factor of <math>n!</math>. Also, <math>(n-3)(n-1)=n^2-4n+3</math> is a factor of <math>n!</math>. Since <math>2n<n^2-4n+3</math> for all <math>n\geq8</math>, we can use Bertrand's Postulate again to show that there is at least one prime number <math>p</math> such that <math>n<p<n^2-4n+3</math>. Since we have two consecutive factors of <math>n!</math> and a prime number between the smaller of these factors and <math>n</math>, the condition will not be satisfied for all even <math>n</math> greater than or equal to <math>8</math>.

Revision as of 01:57, 11 April 2024

Find all integers $n \geq 3$ such that the following property holds: if we list the divisors of $n !$ in increasing order as $1=d_1<d_2<\cdots<d_k=n!$, then we have \[d_2-d_1 \leq d_3-d_2 \leq \cdots \leq d_k-d_{k-1} .\]

Solution (Explanation of Video)

We can start by verifying that $n=3$ and $n=4$ work by listing out the factors of $3!$ and $4!$. We can also see that $n=5$ does not work because the terms $15, 20$, and $24$ are consecutive factors of $5!$. Also, $n=6$ does not work because the terms $6, 8$, and $9$ appear consecutively in the factors of $6!$.

Note that if we have a prime number $p>n$ and an integer $k>p$ such that both $k$ and $k+1$ are factors of $n!$, then the condition cannot be satisfied.

If $n\geq7$ is odd, then $(2)(\frac{n-1}{2})(n-1)=n^2-2n+1$ is a factor of $n!$. Also, $(n-2)(n)=n^2-2n$ is a factor of $n!$. Since $2n<n^2-2n$ for all $n\geq7$, we can use Bertrand's Postulate to show that there is at least one prime number $p$ such that $n<p<n^2-2n$. Since we have two consecutive factors of $n!$ and a prime number between the smaller of these factors and $n$, the condition will not be satisfied for all odd $n$ greater than or equal to $7$.

If $n\geq8$ is even, then $(2)(\frac{n-2}{2})(n-2)=n^2-4n+4$ is a factor of $n!$. Also, $(n-3)(n-1)=n^2-4n+3$ is a factor of $n!$. Since $2n<n^2-4n+3$ for all $n\geq8$, we can use Bertrand's Postulate again to show that there is at least one prime number $p$ such that $n<p<n^2-4n+3$. Since we have two consecutive factors of $n!$ and a prime number between the smaller of these factors and $n$, the condition will not be satisfied for all even $n$ greater than or equal to $8$.

Therefore, the only numbers that work are $n=3$ and $n=4$.

~alexanderruan

Video Solution

https://youtu.be/ZcdBpaLC5p0 [video contains problem 1 and problem 4]

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