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If n = 4p, where p is a prime number greater than 2, how man

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

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

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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


Since we cannot have two correct answers just pick a prime greater than 2, and see how many different positive even divisors will 4p have.

p = 3 --> 4p = 12--> 12 has 4 even divisors: 2, 4, 6, and 12.

Answer: C.

Or this way: since p is prime greater than 2, then p=odd, thus 4p=even, which means that it has 4 even divisors: 2, 4, 2p, and 4p.

Similar question to practice: if-n-is-a-prime-number-greater-than-3-what-is-the-remainder-137869.html

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

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Bunuel wrote:
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


Since we cannot have two correct answers just pick a prime greater than 2, and see how many different positive even divisors will 4p have.

p = 3 --> 4p = 12--> 12 has 4 even divisors: 2, 4, 6, and 12.

Answer: C.

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.

Similar question to practice: if-n-is-a-prime-number-greater-than-3-what-is-the-remainder-137869.html

Hope it helps.


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

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New post 13 Apr 2013, 03:49
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Dipankar6435 wrote:
Bunuel wrote:
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


Since we cannot have two correct answers just pick a prime greater than 2, and see how many different positive even divisors will 4p have.

p = 3 --> 4p = 12--> 12 has 4 even divisors: 2, 4, 6, and 12.

Answer: C.

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.

Similar question to practice: if-n-is-a-prime-number-greater-than-3-what-is-the-remainder-137869.html

Hope it helps.


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


Typo edited. Thank you. +1.
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]

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New post 21 Apr 2013, 00:04
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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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]

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New post 04 Mar 2014, 03:45
p can get any prime number greater than 2 so there should be unlimited different divisor for the n because p can be 11,13,17,19....

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

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lool 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

p can get any prime number greater than 2 so there should be unlimited different divisor for the n because p can be 11,13,17,19....

What is wrong with this ?


No matter which prime p is, 4p will have only four EVEN divisors: 2, 4, 2p, and 4p. Try to check it with any prime greater than 2.

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

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New post 04 Mar 2014, 21:19
lool wrote:
p can get any prime number greater than 2 so there should be unlimited different divisor for the n because p can be 11,13,17,19....

What is wrong with this ?



4 has two even divisors >> 2 & 4

Any prime no p divisors will have 2p & 4p (from above)

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

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

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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, 01:59, edited 1 time in total.
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink]

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New post 02 Jun 2014, 00:04
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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

Hi Bunuel, could you validate my logic pls?


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

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

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New post 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

Hi Bunuel, could you validate my logic pls?



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

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New post 03 Sep 2015, 22:47
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Hi All,

This question can be solved rather easily by TESTing VALUES:

We're told that N = 4P and that P is a prime number greater than 2. Let's TEST P = 3; so N = 12

The question now asks how many DIFFERENT positive EVEN divisors does 12 have, including 12?

12:
1,12
2,6
3,4

How many of these divisors are EVEN? 2, 4, 6, and 12 …..that's a total of 4 even divisors.

Final Answer:
[Reveal] Spoiler:
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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New post 26 Oct 2015, 03:07
n = 4p
p prime no. \(> 2\)

We have to find the no. of even divisors which means even factors of n

P must be a odd no. because 2 is the only even prime no.

Let p be \(3\)

\(n= 4 * 3\)

= \(2^{2} * 3\)

Now , No of factors of P = \(2^{2+1} * 3^{1+1}\) = \(2^{3} * 3^{2}\)

= \(3 * 2 = 6\)

Even factors = Total factors - odd factors

To find Odd factors we take all the prime apart from 2. so here we are left with only 3

Odd factors = \(3^{1+1} = 3^{2}= 2\)

Total factors - Odd factors = Even factors
\(6-2 = 4\)

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

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Re: If n = 4p, where p is a prime number greater than 2, how man   [#permalink] 22 Jan 2016, 10:03
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