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Revision as of 09:52, 18 April 2024
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)
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
Video Solution
https://youtu.be/ZcdBpaLC5p0 [video contains problem 1 and problem 4]
See Also
2024 USAMO (Problems • Resources) | ||
Preceded by Problem 0 |
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.