Special Highly Composite Number/Examples/2520

== Example of Special Highly Composite Number ==
$2520$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.
== Proof ==
By inspection of the sequence of highly composite numbers, $2520$ is highly composite.
For reference, the prime decomposition of $2520$ is:
:$2520 = 2^3 \times 3^2 \times 5 \times 7$
 $n > 2520$ is a highly composite number which is not divisible by $2520$.
We have that $60$ is a special highly composite number.
Therefore $60$ is a divisor of $n$.
It follows that $3$, $4$ and $5$ are all divisors of $n$.
But as $2520$ is not a divisor of $n$, it follows that at least one of $7$, $8$ and $9$ is not a divisor of $n$.
These will be investigated on a case-by-case basis.
:$(1): \quad 7$ is not a divisor of $n$.
By Prime Decomposition of Highly Composite Number we have that:
:$n = 2^a \times 3^b \times 5^c$
where $a \ge b \ge c \ge 1$.
Suppose that $a < 3$.
Then:
:$n \le 2^2 \times 3^2 \times 5^2 = 900$
which is too small.
So we have that $a \ge 3$.
Then:
Now suppose that $b < 2$.
Then:
:$n \le 2^5 \times 3 \times 5 = 480$
which is too small.
So we have that $b \ge 2$.
Then:
But $c \le b$ and so $c < 3$ as well.
Thus we have upper bounds on $a$, $b$ and $c$.
Since $2^a \times 3^b \times 5^c > 2520$, it must be the case that:
:$n = 2^4 \times 3^2 \times 5^2$
which gives that $n = 3600$.
But:
:from we have that $\map {\sigma_0} {3600} = 45$
:from we have that $\map {\sigma_0} {2520} = 48$
This contradicts our hypothesis that $3600$ is highly composite.
By Proof by Contradiction it follows that $7$ must be a divisor of $n$.
:$(2): \quad 9$ is not a divisor of $n$, but $7$ is.
By Prime Decomposition of Highly Composite Number we have that:
:$n = 2^a \times 3^1 \times 5^1 \times 7^1 \times 11^e \times r$
where:
:$e$ is either $0$ or $1$
:$r$ is a possibly vacuous square-free product of prime numbers strictly greater than $11$.
Suppose $e = 1$.
Then:
So $e = 0$ and so by Prime Decomposition of Highly Composite Number $r = 1$.
Thus:
:$n = 2^a \times 3 \times 5 \times 7$
We have that $n > 2520$, so:
:$(2 \text a): a \ge 5$
Then:
It follows by Proof by Contradiction that $9$ is a divisor of $n$.
:$(3): \quad 8$ is not a divisor of $n$, but $7$ and $9$ both are.
By Prime Decomposition of Highly Composite Number we have that:
:$n = 2^2 \times 3^2 \times 5^c \times 7^d \times 11^e \times r$
where $r$ is a possibly vacuous product of prime numbers strictly greater than $11$.
Suppose:
:$(3 \text a): \quad e > 0$
Then:
So $e = 0$ and so Prime Decomposition of Highly Composite Number $r = 1$.
Thus:
:$n = 2^2 \times 3^2 \times 5^c \times 7^d$
Suppose $c = 2$.
Then:
The remaining possibility is that $c = 1$ and $d = 1$:
Thus:
:$n = 2^2 \times 3^2 \times 5 \times 7 = 1260$
But this is a contradiction of our supposition that $n > 2520$.
It follows by Proof by Contradiction that $8$ is a divisor of $n$.
By Proof by Cases it is seen that the existence of a highly composite $n$ not divisible by $2520$ leads to a contradiction.
The result then follows by Proof by Contradiction.
== Sources ==
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Category:Special Highly Composite Numbers
Category:2520