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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]

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01 Jun 2014, 22:06

3

This post received KUDOS

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n= 4p = 2^2*p^1 so total no. of factors: (2+1)*(1+1)= 6 total no. of odd factors, since p is odd as it is a prime>2: p^1 and p^0 = 2 Total no. of even factors: 6 - 2 = 4

Now if n was n=4pq where p and q are both prime no.s greater than 2 then: total no. of factors: (2+1)*(1+1)*(1+1)= 12 total no. of odd factors, since p is odd as it is a prime>2: (1+1)*(1+1)= 4 Total no. of even factors: 12 - 4 = 8

Hi Bunuel, could you validate my logic pls? _________________

Please consider giving 'kudos' if you like my post and want to thank

Hi Bunnel Just a small typo in the explanation Or this way: since p is prime greater than 2, then p=odd, thus 4p=odd, which means that it has 4 even divisors: 2, 4, 2p, and 4p. It should be 4p = Even Cheers

Hi Bunnel Just a small typo in the explanation Or this way: since p is prime greater than 2, then p=odd, thus 4p=odd, which means that it has 4 even divisors: 2, 4, 2p, and 4p. It should be 4p = Even Cheers

Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]

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21 Apr 2013, 00:04

Is one not a divisor? _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]

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19 Apr 2014, 23:35

n=4*p

And p(prime) > 2

n= 2*2*p

Positive even divisors 'n' can have including 'n':

2 2p 4 4p which is also equal to 'n'

Hence 4 positive even divisors. _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]

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02 Jun 2014, 00:04

Expert's post

MensaNumber wrote:

n= 4p = 2^2*p^1 so total no. of factors: (2+1)*(1+1)= 6 total no. of odd factors, since p is odd as it is a prime>2: p^1 and p^0 = 2 Total no. of even factors: 6 - 2 = 4

Now if n was n=4pq where p and q are both prime no.s greater than 2 then: total no. of factors: (2+1)*(1+1)*(1+1)= 12 total no. of odd factors, since p is odd as it is a prime>2: (1+1)*(1+1)= 4 Total no. of even factors: 12 - 4 = 8

Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]

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17 Apr 2015, 02:36

if we apply theory of finding number of divisor Let,s plug in p= 3 then n= 12 Factor of 12 are 3, 2, 2 p>2, so 2s are out Now the formula (p+1)+(q+1)+(r+1) = 3+1 = 4 Ans C _________________

Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]

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01 Sep 2015, 03:30

MensaNumber wrote:

n= 4p = 2^2*p^1 so total no. of factors: (2+1)*(1+1)= 6 total no. of odd factors, since p is odd as it is a prime>2: p^1 and p^0 = 2 Total no. of even factors: 6 - 2 = 4

Now if n was n=4pq where p and q are both prime no.s greater than 2 then: total no. of factors: (2+1)*(1+1)*(1+1)= 12 total no. of odd factors, since p is odd as it is a prime>2: (1+1)*(1+1)= 4 Total no. of even factors: 12 - 4 = 8

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