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)

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

Video Solution

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

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2024 USAMO Problems/Problem 1

Solution (Explanation of Video)

Video Solution