2024 USAMO Problems/Problem 1

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 Solution)

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 $n\geq7$.

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 $n\geq8$.

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

~alexanderruan

As listed above, $n=3$ and $n=4$ work but $n=5$ and $n=6$ don't. Assume $n>6$, let $p$ be the smallest positive integer that doesn't divide $n!$. It is easy to see $p$ is the smallest prime greater than $n$. It is easy to see that both $p-1$ and $p+1$ are divisors of $n!$ Let $q$ be the smallest positive integer greater than $p$ that is 2 mod 6. We see that either $q=p+3$ or $q=p+1$ but $q$, $q+1$, $q+2$ must all be divisors of $n!$ giving a difference of 1 after a difference of 2. Therefore, all $n>6$ fail. ~mathophobia

Solution 2

We claim only $n = 3$ and $n = 4$ are the only two solutions. First, it is clear that both solutions work.

Next, we claim that $n < 5$. For $n \geq 5$, let $x$ be the smallest $x$ such that $x+1$ is not a factor of $n!$. Let the smallest factor larger than $x$ be $x+k$.

Now we consider $\frac{n!}{x-1}$, $\frac{n!}{x}$ and $\frac{n!}{x+k}$. Since $\frac{n!}{x-1} > \frac{n!}{x} > \frac{n!}{x+k}$, if $n$ were to satisfy the conditions, then $\frac{n!}{x-1}-\frac{n!}{x} \geq \frac{n!}{x} - \frac{n!}{x+k}$. However, note that this is not true for $x \geq 5$ and $k > 1$.

Note that the inequality is equivalent to showing $\frac{1}{x(x-1)} \geq \frac{k}{x(x+k)}$, which simplifies to $x+k \geq kx-k$, or $\frac{x}{x-2} \geq k \geq 2$. This implies $x \geq 2x-4$, $x \leq 4$, a contradiction, since the set of numbers $\{1, 2, 3, 4, 5\}$ are all factors of $n!$, and the value of $x$ must exist. Hence, no solutions for $n \geq 5$.

Video Solution

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

See Also

2024 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png