Special Highly Composite Number/Examples/60
== Example of Special Highly Composite Number ==  $60$ 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, $60$ is highly composite.  $n > 60$ is a highly composite number which is not divisible by $60$. We have that $12$ is a special highly composite number. Therefore $12$ is a divisor of $n$. As $60$ is not a divisor of $n$, it follows that while $3$ is a divisor of $n$, $5$ is not. From Prime Decomposition of Highly Composite Number, no prime number greater than $5$ is a divisor of $n$. Thus: :$n = 2^a \times 3^b$ where $a \ge b \ge 1$. This will be investigated on a case-by-case basis. :$(1): \quad b = 1$ That is, $n = 2^a \times 3$. We have that $n > 60$. Therefore: :$(1 \text a): \quad a \ge 5$ as $2^4 \times 3^1 = 48$. Then: {{begin-eqn}} {{eqn | l = 2^{a - 3} \times 3 \times 5       | o = <       | r = 2^a \times 3       | c = because $5 < 2^3 = 8$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 3} \times 3 \times 5}\right)       | o = <       | r = \tau \left({2^a \times 3}\right)       | c = as $2^a \times 3$ is highly composite }} {{eqn | ll= \leadsto       | l = \left({\left({a - 3}\right) + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({a + 1}\right) \left({1 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 4 \left({a - 2}\right)       | o = <       | r = 2 \left({a + 1}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = a       | o = <       | r = 5       | c = which is a contradiction of $(1 \text a)$ }} {{end-eqn}} It follows by Proof by Contradiction that $b \ne 1$. {{qed|lemma}} :$(2): \quad b = 2$ That is, $n = 2^a \times 3^2$. We have that $n > 60$. Therefore: :$(2 \text a): \quad a \ge 3$ as $2^2 \times 3^2 = 36$. Then: {{begin-eqn}} {{eqn | l = 2^{a - 1} \times 3 \times 5       | o = <       | r = 2^a \times 3^2       | c = because $5 < 2 \times 3 = 6$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 1} \times 3 \times 5}\right)       | o = <       | r = \tau \left({2^a \times 3^2}\right)       | c = as $2^a \times 3^2$ is highly composite }} {{eqn | ll= \leadsto       | l = \left({\left({a - 1}\right) + 1}\right) \left({1 + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({a + 1}\right) \left({2 + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 4 a       | o = <       | r = 3 \left({a + 1}\right)       | c = simplifying }} {{eqn | ll= \leadsto       | l = a       | o = <       | r = 3       | c = which is a contradiction of $(2 \text a)$ }} {{end-eqn}} It follows by Proof by Contradiction that $b \ne 2$. {{qed|lemma}} :$(3): \quad b \ge 3$ By Prime Decomposition of Highly Composite Number we have that $a \ge 3$. Then: {{begin-eqn}} {{eqn | l = 2^{a - 1} \times 3^{b - 1} \times 5       | o = <       | r = 2^a \times 3^b       | c = because $5 < 2 \times 3 = 6$ }} {{eqn | ll= \leadsto       | l = \tau \left({2^{a - 1} \times 3^{b - 1} \times 5}\right)       | o = <       | r = \tau \left({2^a \times 3^b}\right)       | c = as $2^a \times 3^b$ is highly composite }} {{eqn | ll= \leadsto       | l = \left({\left({a - 1}\right) + 1}\right) \left({\left({b - 1}\right) + 1}\right) \left({1 + 1}\right)       | o = <       | r = \left({a + 1}\right) \left({b + 1}\right)       | c = {{Defof|Tau Function}} }} {{eqn | ll= \leadsto       | l = 2 a b       | o = <       | r = \left({a + 1}\right) \left({b + 1}\right)       | c =  }} {{eqn | ll= \leadsto       | l = a b       | o = <       | r = a + b + 1       | c =  }} {{eqn | ll= \leadsto       | l = 3 a       | o = <       | r = a + b + 1       | c = as $b \ge 3$ }} {{eqn | ll= \leadsto       | l = 2 a       | o = <       | r = b + 1       | c =  }} {{eqn | ll= \leadsto       | l = 2 b       | o = <       | r = b + 1       | c = as $a \ge b$ }} {{eqn | ll= \leadsto       | l = b       | o = <       | r = 1       | c = which is a contradiction of $(3)$ }} {{end-eqn}} {{qed|lemma}} By Proof by Cases it is seen that the existence of a highly composite $n$ not divisible by $60$ leads to a contradiction. The result then follows by Proof by Contradiction.  == Sources == * {{Citation|title = An Interesting Subset of the Highly Composite Numbers|author = Steven Ratering|journal = Mathematics Magazine|abbr = Math. Mag.|volume = 64|issue = 5|date = Dec. 1991|startpage = 343|endpage = 346|jstor = 2690653}}