# Special Highly Composite Number/Examples/12

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## 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.

$\blacksquare$

## Sources

- Dec. 1991: Steven Ratering:
*An Interesting Subset of the Highly Composite Numbers*(*Math. Mag.***Vol. 64**,*no. 5*: pp. 343 – 346) www.jstor.org/stable/2690653