Difference between revisions of "2024 USAMO Problems/Problem 1"
Jeffersonj (talk | contribs) (→See Also) |
|||
(5 intermediate revisions by 2 users not shown) | |||
Line 10: | Line 10: | ||
Note that if we have a prime number <math>p>n</math> and an integer <math>k>p</math> such that both <math>k</math> and <math>k+1</math> are factors of <math>n!</math>, then the condition cannot be satisfied. | Note that if we have a prime number <math>p>n</math> and an integer <math>k>p</math> such that both <math>k</math> and <math>k+1</math> are factors of <math>n!</math>, then the condition cannot be satisfied. | ||
− | If <math>n\geq7</math> is odd, then <math>(2)(\frac{n-1}{2})(n-1)=n^2-2n+1</math> is a factor of <math>n!</math>. Also, <math>(n-2)(n)=n^2-2n</math> is a factor of <math>n!</math>. Since <math>2n<n^2-2n</math> for all <math>n\geq7</math>, we can use [[Bertrand's Postulate]] to show that there is at least one prime number <math>p</math> such that <math>n<p<n^2-2n</math>. Since we have two consecutive factors of <math>n!</math> and a prime number between the smaller of these factors and <math>n</math>, the condition will not be satisfied for all odd <math>n | + | If <math>n\geq7</math> is odd, then <math>(2)(\frac{n-1}{2})(n-1)=n^2-2n+1</math> is a factor of <math>n!</math>. Also, <math>(n-2)(n)=n^2-2n</math> is a factor of <math>n!</math>. Since <math>2n<n^2-2n</math> for all <math>n\geq7</math>, we can use [[Bertrand's Postulate]] to show that there is at least one prime number <math>p</math> such that <math>n<p<n^2-2n</math>. Since we have two consecutive factors of <math>n!</math> and a prime number between the smaller of these factors and <math>n</math>, the condition will not be satisfied for all odd <math>n\geq7</math>. |
− | If <math>n\geq8</math> is even, then <math>(2)(\frac{n-2}{2})(n-2)=n^2-4n+4</math> is a factor of <math>n!</math>. Also, <math>(n-3)(n-1)=n^2-4n+3</math> is a factor of <math>n!</math>. Since <math>2n<n^2-4n+3</math> for all <math>n\geq8</math>, we can use Bertrand's Postulate again to show that there is at least one prime number <math>p</math> such that <math>n<p<n^2-4n+3</math>. Since we have two consecutive factors of <math>n!</math> and a prime number between the smaller of these factors and <math>n</math>, the condition will not be satisfied for all even <math>n | + | If <math>n\geq8</math> is even, then <math>(2)(\frac{n-2}{2})(n-2)=n^2-4n+4</math> is a factor of <math>n!</math>. Also, <math>(n-3)(n-1)=n^2-4n+3</math> is a factor of <math>n!</math>. Since <math>2n<n^2-4n+3</math> for all <math>n\geq8</math>, we can use Bertrand's Postulate again to show that there is at least one prime number <math>p</math> such that <math>n<p<n^2-4n+3</math>. Since we have two consecutive factors of <math>n!</math> and a prime number between the smaller of these factors and <math>n</math>, the condition will not be satisfied for all even <math>n\geq8</math>. |
Therefore, the only numbers that work are <math>n=3</math> and <math>n=4</math>. | Therefore, the only numbers that work are <math>n=3</math> and <math>n=4</math>. | ||
Line 20: | Line 20: | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/ZcdBpaLC5p0 [video contains problem 1 and problem 4] | https://youtu.be/ZcdBpaLC5p0 [video contains problem 1 and problem 4] | ||
+ | |||
+ | ==See Also== | ||
+ | {{USAMO newbox|year=2024|before=First Question|num-a=2}} | ||
+ | {{MAA Notice}} |
Revision as of 10:54, 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 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.