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

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

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New post 27 Dec 2012, 04:55
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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
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Re: If n = 4p, where p is a prime number greater than 2, how many differen  [#permalink]

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New post 27 Dec 2012, 04:59
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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 many differen  [#permalink]

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New post Updated on: 02 Jun 2014, 00:59
6
4
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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Re: If n = 4p, where p is a prime number greater than 2, how many differen  [#permalink]

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New post 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 many differen  [#permalink]

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

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New post 04 Mar 2014, 02:45
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 many differen  [#permalink]

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New post 04 Mar 2014, 02:53
1
2
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 many differen  [#permalink]

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New post 04 Mar 2014, 20:19
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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 many differen  [#permalink]

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New post 19 Apr 2014, 22:35
1
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 many differen  [#permalink]

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New post 01 Jun 2014, 23:04
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 many differen  [#permalink]

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New post 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 many differen  [#permalink]

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New post 21 Aug 2015, 10:16
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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, 12 …..4 even divisors.

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

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New post 03 Sep 2015, 21:47
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:

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

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New post 26 Oct 2015, 02:07
1
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 many differen  [#permalink]

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

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New post 01 Jun 2016, 07:47
Since p is primes and is > 2, then p must be an odd number since 2 is the only even prime number.

I solved by simply plugging in:
p = 5 (a prime) ---> n = 20
Factors of 20 include:
20 1
10 2
5 4

In this factor tree there is a total of 4 even factors means C is the correct answer.

If you aren't convinced after this you should have time to plug in another prime number for p. I finished within 50 seconds using just p =5.
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Re: If n = 4p, where p is a prime number greater than 2, how many differen  [#permalink]

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New post 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 many differen  [#permalink]

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New post 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

Hence C

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

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New post 01 Apr 2017, 06:09
n=4p
Let;s factorize n
n= 2^2 * p^a (a is some positive power of p and we know that p is a prime number greater than 2 i.e. it is odd as the only even prime number is 2. So we don't have to worry about the power of p as it would yield odd integer and we have to just find even factors.)

Thus the answer is 2, 4(2^2), 2p (even), and 4p (even).
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Re: If n = 4p, where p is a prime number greater than 2, how many differen  [#permalink]

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New post 26 Sep 2018, 21:36
Would this be an acceptable way of solving:

Prime factor the 4 that is next to p
Prime factorization gives us 2^1 and 2^1
Using only Even primes with above Powers of 1, we just add 1 to the power of each and multiply - this would give us different combinations of even factors
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Re: If n = 4p, where p is a prime number greater than 2, how many differen &nbs [#permalink] 26 Sep 2018, 21:36
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