Mixing
The numbers of the even divisors and the odd divisors of a natural number
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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
elementary-number-theory
asked Jan 30 at 15:41
Joshua Haim MamouJoshua Haim Mamou
245
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add a comment |
$begingroup$
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
elementary-number-theory
asked Jan 30 at 15:41
Joshua Haim MamouJoshua Haim Mamou
245
$endgroup$
add a comment |
$begingroup$
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
elementary-number-theory
asked Jan 30 at 15:41
Joshua Haim MamouJoshua Haim Mamou
245
$endgroup$
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
elementary-number-theory
elementary-number-theory
asked Jan 30 at 15:41
Joshua Haim MamouJoshua Haim Mamou
245
asked Jan 30 at 15:41
Joshua Haim MamouJoshua Haim Mamou
245
asked Jan 30 at 15:41
Joshua Haim MamouJoshua Haim Mamou
245
asked Jan 30 at 15:41
Joshua Haim MamouJoshua Haim Mamou
245
asked Jan 30 at 15:41
Joshua Haim MamouJoshua Haim Mamou
245
245
add a comment |
add a comment |
1 Answer
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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.
answered Jan 30 at 15:45
ArthurArthur
122k7122210
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How do u prove that fact?
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– 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
add a comment |
Your Answer
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1 Answer
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$begingroup$
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.
answered Jan 30 at 15:45
ArthurArthur
122k7122210
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 30 at 15:45
ArthurArthur
122k7122210
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 30 at 15:45
ArthurArthur
122k7122210
$endgroup$
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.
answered Jan 30 at 15:45
ArthurArthur
122k7122210
edited Jan 30 at 15:54
answered Jan 30 at 15:45
ArthurArthur
122k7122210
answered Jan 30 at 15:45
ArthurArthur
122k7122210
answered Jan 30 at 15:45
ArthurArthur
122k7122210
122k7122210
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
How do u prove that fact?
$endgroup$
– Joshua Haim Mamou
Jan 30 at 16:21
$begingroup$
How do u prove that fact?
$endgroup$
– Joshua Haim Mamou
Jan 30 at 16:21
1
1
$begingroup$
@JoshuaHaimMamou My example is basically a full proof. It just needs some minor adjustments.
$endgroup$
– Arthur
Jan 30 at 16:22
$begingroup$
@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
$begingroup$
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
$begingroup$
@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
add a comment |
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