Difference between revisions of "2024 USAMO Problems/Problem 1"
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d_2-d_1 \leq d_3-d_2 \leq \cdots \leq d_k-d_{k-1} . | d_2-d_1 \leq d_3-d_2 \leq \cdots \leq d_k-d_{k-1} . | ||
</cmath> | </cmath> | ||
+ | |||
+ | ==Solution (Explanation of Video)== | ||
+ | |||
+ | We can start by verifying that <math>n=3</math> and <math>n=4</math> work by listing out the factors of <math>3!</math> and <math>4!</math>. We can also see that <math>n=5</math> does not work because the terms <math>15, 20</math>, and <math>24</math> are consecutive factors of <math>5!</math>. Also, <math>n=6</math> does not work because the terms <math>6, 8</math>, and <math>9</math> appear consecutively in the factors of <math>6!</math>. | ||
+ | |||
+ | 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 odd <math>n</math> greater than or equal to <math>7</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 even <math>n</math> greater than or equal to <math>8</math>. | ||
+ | |||
+ | Therefore, the only numbers that work are <math>n=3</math> and <math>n=4</math>. | ||
+ | |||
+ | ~alexanderruan | ||
==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] |
Revision as of 01:52, 11 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 odd
greater than or equal to
.
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 even
greater than or equal to
.
Therefore, the only numbers that work are and
.
~alexanderruan
Video Solution
https://youtu.be/ZcdBpaLC5p0 [video contains problem 1 and problem 4]