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:

{{eqn | ll= \leadsto | l = \map {\sigma_0} {2^4 \times 5} \times \map {\sigma_0} {3^2 \times 7^d} | o = < | r = \map {\sigma_0} {2^2 \times 5^2} \times \map {\sigma_0} {3^2 \times 7^d | c = Divisor Counting Function is Multiplicative }}

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.