2024 USAMO Problems/Problem 1
Find all integers such that the following property holds: if we list the divisors of in increasing order as , then we have
Solution (Explanation of Video Solution)
We can start by verifying that and work by listing out the factors of and . We can also see that does not work because the terms , and are consecutive factors of . Also, does not work because the terms , and appear consecutively in the factors of .
Note that if we have a prime number and an integer such that both and are factors of , then the condition cannot be satisfied.
If is odd, then is a factor of . Also, is a factor of . Since for all , we can use Bertrand's Postulate to show that there is at least one prime number such that . Since we have two consecutive factors of and a prime number between the smaller of these factors and , the condition will not be satisfied for all odd .
If is even, then is a factor of . Also, is a factor of . Since for all , we can use Bertrand's Postulate again to show that there is at least one prime number such that . Since we have two consecutive factors of and a prime number between the smaller of these factors and , the condition will not be satisfied for all even .
Therefore, the only numbers that work are and .
~alexanderruan
As listed above, and work but and don't. Assume , let be the smallest positive integer that doesn't divide . It is easy to see is the smallest prime greater than . It is easy to see that both and are divisors of Let be the smallest positive integer greater than that is 2 mod 6. We see that either or but , , must all be divisors of giving a difference of 1 after a difference of 2. Therefore, all fail. ~mathophobia
Solution 2
We claim only and are the only two solutions. First, it is clear that both solutions work.
Next, we claim that . For , let be the smallest such that is not a factor of . Let the smallest factor larger than be .
Now we consider , and . Since , if were to satisfy the conditions, then . However, note that this is not true for and .
Note that the inequality is equivalent to showing , which simplifies to , or . This implies , , a contradiction, since the set of numbers are all factors of , and the value of must exist. Hence, no solutions for .
Video Solution
https://youtu.be/ZcdBpaLC5p0 [video contains problem 1 and problem 4]
See Also
2024 USAMO (Problems • Resources) | ||
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.