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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, 21:06

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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?
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Last edited by NoHalfMeasures on 02 Jun 2014, 00:59, edited 1 time in total.

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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20 Apr 2013, 23:04

Is one not a divisor?
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]

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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.
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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

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

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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
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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 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

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

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13 Jul 2016, 06:39

Walkabout wrote:

If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) Two (B) Three (C) Four (D) Six (E) Eight

This is an interesting question because we are immediately given the option to insert any prime number we wish for p. Since this is a problem-solving question, and there can only be one correct answer, we can select any value for p, as long as it is a prime number greater than 2. We always want to work with small numbers, so we should select 3 for p. Thus, we have:

n = 4 x 3

n = 12

Next we have to determine all the factors, or divisors, of P. Remember the term factor is synonymous with the term divisor.

1, 12, 6, 2, 4, 3

From this we see that we have 4 even divisors: 12, 6, 2, and 4.

If you are concerned that trying just one value of p might not substantiate the answer, try another value for p. Let’s say p = 5, so

n = 4 x 5

n = 20

The divisors of 20 are: 1, 20, 2, 10, 4, 5. Of these, 4 are even: 20, 2, 10 and 4. As we can see, again we have 4 even divisors.

No matter what the value of p, as long as it is a prime number greater than 2, n will always have 4 even divisors.

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

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28 Dec 2016, 07:16

One of those Quality Questions from the Official Guide. here is my solution => Method 1-> n=2^2*p as p is a odd prime (all primes >2 are odd) number of even factors => 2*2-> four Four factors are -> 2,4,2p,4p Alternatively let p=3 so n=12 factors => 1,2,3,4,6,12 even factors => 2,4,6,12 hence four

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