# Special Highly Composite Number/Examples/12

## Example of Special Highly Composite Number

$12$ 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, $12$ is highly composite.

Aiming for a contradiction, suppose $n > 12$ is a highly composite number which is not divisible by $12$.

We have that $6$ is a special highly composite number.

Therefore $6$ is a divisor of $n$.

As $12$ is not a divisor of $n$, it follows that the multiplicity of $2$ in $n$ is $1$.

From Prime Decomposition of Highly Composite Number, that means:

$n = 2 \times 3 \times 5 \times r$

where $r$ is a possibly vacuous square-free product of prime numbers strictly greater than $5$.

Then:

 $\ds 2^3 \times 3 \times r$ $<$ $\ds 2 \times 3 \times 5 \times r$ because $4 = 2^2 < 5$ $\ds \leadsto \ \$ $\ds \map {\sigma_0} {2^3 \times 3 \times r}$ $<$ $\ds \map {\sigma_0} {2 \times 3 \times 5 \times r}$ as $2 \times 3 \times 5 \times r$ is highly composite $\ds \leadsto \ \$ $\ds \map {\sigma_0} {2^3 \times 3} \times \map {\sigma_0} r$ $<$ $\ds \map {\sigma_0} {2 \times 3 \times 5} \times \map {\sigma_0} r$ Divisor Count Function is Multiplicative $\ds \leadsto \ \$ $\ds \map {\sigma_0} {2^3 \times 3}$ $<$ $\ds \map {\sigma_0} {2 \times 3 \times 5}$ simplifying $\ds \leadsto \ \$ $\ds \paren {3 + 1} \paren {1 + 1}$ $<$ $\ds \paren {1 + 1} \paren {1 + 1} \paren {1 + 1}$ Definition of Divisor Count Function $\ds \leadsto \ \$ $\ds 8$ $<$ $\ds 8$ which is a falsehood

The result follows by Proof by Contradiction.

$\blacksquare$