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This question is in reference to this answer to a question I asked.

It can be proved that numbers that have most factors than the numbers before it have the degrees of their prime factors arranged in the descending order-:

$N = 2^p\cdot 3^q\cdot5^r\cdot\ldots \text{ so on}$

where

$p \ge q \ge r$

I want to know why this is the case.

What is the intuition behind it?

What I can know is that the lowest prime factors will occur more and more as I move towards $\infty+$. But it is not clear to me why this property is followed by the number that has the maximum factors.

For example:

Take $2^2\cdot3^3$ and $2^3\cdot3^4$ they have a total number of factors of $12$ and $20$, yet the second number does not follow the above-mentioned property.

Also if someone can point me to a proof of this, it would be great.

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    $\begingroup$ If this were not the case, you could get a smaller number with the same number of divisors by sorting the exponents. More details : en.wikipedia.org/wiki/Highly_composite_number $\endgroup$ – Peter Apr 6 '18 at 8:28
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    $\begingroup$ The intuition is, the number of factors depends on exponents only, not their order. So assigning larger exponents to small primes rather than the large ones will give you a smaller number. $\endgroup$ – Wojowu Apr 6 '18 at 8:28
  • $\begingroup$ @Wojowu You might want to make your comment an answer. $\endgroup$ – ng.newbie Apr 6 '18 at 9:06
  • $\begingroup$ @Peter You might want to make your comment into an answer. $\endgroup$ – ng.newbie Apr 6 '18 at 9:07
  • $\begingroup$ Sadly right now I don't have the time. If anyone is willing to write up an answer, feel free to :) $\endgroup$ – Wojowu Apr 6 '18 at 9:07
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As Wojowu mentioned, the number of divisors only depends on the exponents in the prime factorization, to be more precise, if $$N=p_1^{a_1}\cdots p_n^{a_n}$$ then the number of divisors is $$(a_1+1)\cdots (a_n+1)$$ Hence changing the order of the exponents does not change the number of divisors. But the smallest number we have with a given set of exponents is $2^{a_1}\cdot 3^{a_2}\cdot 5^{a_3}\cdots p_n^{a_n}$ with $a_1\ge a_2\ge \cdots \ge a_n$ if $p_n$ is the $n$-th prime.

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