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]

2024 USAMO (Problems • Resources)
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2024 USAMO Problems/Problem 1

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Solution (Explanation of Video Solution)

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