# Why does the with the most divisors than numbers smaller than it, have the degree of the prime number ordered in descending order?

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.

• 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 – Peter Apr 6 '18 at 8:28
• 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. – Wojowu Apr 6 '18 at 8:28
• @Wojowu You might want to make your comment an answer. – ng.newbie Apr 6 '18 at 9:06
• @Peter You might want to make your comment into an answer. – ng.newbie Apr 6 '18 at 9:07
• Sadly right now I don't have the time. If anyone is willing to write up an answer, feel free to :) – Wojowu Apr 6 '18 at 9:07

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.