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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]
20 Apr 2013, 23:04
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]
19 Apr 2014, 22:35
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]
01 Jun 2014, 21:06
2
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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
Last edited by NoHalfMeasures on 02 Jun 2014, 00:59, edited 1 time in total.
Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]
01 Jun 2014, 23:04
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]
17 Apr 2015, 01:36
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]
01 Sep 2015, 02:30
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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