# Number of divisors of n after multiplication

I was trying to find numbers with same number of odd and even divisors. The solution is $2l$ where $l$ is odd and I think I understand this - you're adding one even divisor for every odd one that was there.

If you take power of 2 and multiply by 2 you increase the number of even divisors by 1. This also makes sense to me since you keep all the previous divisors and add one new.

(paragraph below NOT true, you get new even divisor for every odd one, see comment)
But it looks like (I don't know if this is actually true) that if you multiply any even number that is not power of 2 by 2 you increase the number of even divisors by 2 (while the number of odd divisors doesn't change).

So what's happening here? Can this be more generalized (number of divisors with respect to parity)?

It looks to me you get one new divisor - the new number (makes sense) and another one for some other number that doesn't "overlap" (2 -> 4 -> 8 -> 16 -> ... when you start with number 6 and multiply by 2).

I hope this question makes sense but please ask for clarifications if not. It's also possible I'm just looking at numbers for too long.

• The third para is not clear to me – Archis Welankar Dec 30 '15 at 17:32
• Let your "old" number $l$ be $2^km$, where $m$ is odd. Now consider the divisors of $2l$. The "new" divisors added are of the form $2^{k+1}d$, where $d$ ranges over the divisors of $m$. So the number of divisors added is just the number of divisors of $m$. – André Nicolas Dec 30 '15 at 17:46
• Wouldn't 4 have the same number of odd and even divisors at 1 each? – JB King Dec 30 '15 at 18:20
• @ArchisWelankar That paragraph is not true as shown by André. – David Mašek Dec 30 '15 at 19:08
• @JBKing divs(4) = {1, 2, 4} I guess you forgot 4? – David Mašek Dec 30 '15 at 19:09

Let $l=2^{k}m$, where $k\ge 1$, and $m$ is odd. From every odd divisor $d$ of $m$ (and hence of $l$), we can obtain an even divisor of $l$ by multiplying $d$ by one of $2,2^2,\dots,2^k$.
So if $l$ has $x$ odd divisors, then it has $kx$ even divisors.