0
$\begingroup$

Can a natural number have an odd number of the even divisors and an even number of the odd divisors?

Example: All the prime numbers have an even number of the odd divisors (2), but also 0 even divisors (we want odd number of even divisors and 0 is considered as even number)

9 has odd number (3) of odd divisors so it won't fit either

$\endgroup$
5
$\begingroup$

No, it cannot. The total number of divisors (and therefore the number of even divisors) must be a multiple of the number of odd divisors.

Say, for instance, that our number is divisible by 4 but not 8. Take all the odd divisors, and multiply them by $2$. You now have all the even divisors which are not divisible by $4$. Multiply them by $2$ again, and you have all the divisors which are divisible by $4$.

So we have here a partition of all the divisors into three equally-sized parts, one of which is exactly all the odd divisors. So the total number of divisors is three times the number of odd divisors, and the number of even divisors is twice that of odd divisors.

A corresponding argument works no matter how many times $2$ goes into our number.

$\endgroup$
  • $\begingroup$ How do u prove that fact? $\endgroup$ – Joshua Haim Mamou Jan 30 at 16:21
  • 1
    $\begingroup$ @JoshuaHaimMamou My example is basically a full proof. It just needs some minor adjustments. $\endgroup$ – Arthur Jan 30 at 16:22
  • $\begingroup$ I meant, how do you prove that the total number of divisors must be a multiple of the number of the odd divisors? $\endgroup$ – Joshua Haim Mamou Jan 30 at 18:28
  • $\begingroup$ @JoshuaHaimMamou And I meant, that's exactly what my example tells you how to prove. With some minor adjustments. $\endgroup$ – Arthur Jan 30 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.