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Problem Solving - Numbers properties

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by koby_gen » Sun Jan 09, 2011 9:34 am
If n=8p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 8
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by anshumishra » Sun Jan 09, 2011 9:44 am
koby_gen wrote:If n=8p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 8
n = 2^3*p
Number of even factors = (Number of ways to select at least one 2 out of 3)*(Number of ways to select any number of p's out of 1) = (3)*(1 + 1) = 6

D
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by anshumishra » Sun Jan 09, 2011 9:47 am
koby_gen wrote:If n=8p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 8
Another way

n = 2^3*p^1 (p is odd)
No. of even factors= All factors - odd factors= (3+1)*(1+1) - (1+1) = 8 - 2 = 6.
Thanks
Anshu

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by towerSpider » Sun Jan 09, 2011 10:34 am
Anshu, i didnt get your method. Can you explain?
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by anshumishra » Sun Jan 09, 2011 10:47 am
towerSpider wrote:Anshu, i didnt get your method. Can you explain?
Sure.
n=2^3*p^1

All the possible factors :
2^0*p^0
2^1*p^0
2^2*p^0
2^3*p^0
2^0*p^1
2^1*p^1
2^2*p^1
2^3*p^1

Out of these 8 factors, only two (in bold) are odd (As the power of 2 which is multiplied to it is 0).
Try to look at this as combination problem now. You are trying to select all the possible powers of 2 here except when it is 0.

So, it will be :
(Number of ways to select at least one 2 out of 3)*(Number of ways to select any number of p's out of 1) = (3)*(1 + 1) = 6

Hope that clears your doubt.
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by GMATGuruNY » Sun Jan 09, 2011 2:19 pm
koby_gen wrote:If n=8p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 8
We can plug in a value for p.
Let p = 3.
Then n = 8*3 = 24.
Even factors of 24 are 2, 4, 6, 8, 12, 24 = 6 even factors.

The correct answer is D.
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by nipunkathuria » Sun Jan 16, 2011 11:22 am
Would like to add something over here:

N=x^p1*y^q1
where x and y are the prime factors of number N
the number of factors of the number N would be (p1+1)(q1+1)

But in our case the restriction has been that of Even factors.
Hence there will not be any role played by the p(for n=8*p) ...
hence we have 2^3=8 factors - (2^0*p and 2^0 * 1)



hence 6 factors
Back !!!
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by anshumishra » Sun Jan 16, 2011 11:30 am
nipunkathuria wrote:Would like to add something over here:

N=x^p1*y^q1
where x and y are the prime factors of number N
the number of factors of the number N would be (p1+1)(q1+1)

But in our case the restriction has been that of Even factors.
Hence there will not be any role played by the p(for n=8*p) ...
hence we have 2^3=8 factors - (2^0*p and 2^0 * 1)



hence 6 factors
Right, The part in bold is also the number of odd factors, as I have already posted in a post above :
No. of even factors= All factors - odd factors= (3+1)*(1+1) - (1+1) = 8 - 2 = 6.
Thanks
Anshu

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by thebigkats » Thu Jan 20, 2011 11:35 am
calculation in this case seems too complicated.
We know p = odd and prime (means no more factors in p)

this means that only 8 can be broken in factors and we need to find its factors and resulting multiplications from p

So the only even factors can be -

2
4
8
2*p
4*p
8*p

you can't break it any further. so total = 6
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by Everest » Thu Jan 20, 2011 1:28 pm
Prime number greater than 2 are 3, 5, 7, 11.....

just plug in 3 into p, n = 8p

=> n = 24,
=> even divisors are 2, 4, 6 , 8 , 12 and 24.

Therefore there are 6 even integers.
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by sushantgupta » Mon Jul 04, 2011 10:57 am
total 6 divisor

2, 4, 8 , 2p , 4p , 8p
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by gmatblood » Tue Jul 05, 2011 11:13 am
if we substitute a value for p, lets say 4,instead of 3

then,

8*5=40,
positive divisors of 40 are {2,4,5,8,10,20,40} which ends up in 7, which is not the answer !!

so GMATGuruNY and anshumishra are wrong!!
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by Buix0065 » Tue Jul 05, 2011 5:03 pm
Plugging in a prime > 2 for P, 3 would result in 24.

Even divisors of 24 are: 2, 4, 6, 8, 12, n

so I get 6.

Any insights on this methodology? Is there any way that a larger prime number would result in a different outcome?
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by artistocrat » Tue Jul 05, 2011 5:52 pm
Buix0065 wrote:Plugging in a prime > 2 for P, 3 would result in 24.

Even divisors of 24 are: 2, 4, 6, 8, 12, n

so I get 6.

Any insights on this methodology? Is there any way that a larger prime number would result in a different outcome?
There is no other way that a larger prime number would result in a different outcome, because that would violate the question by admitting that there is more than one answer!
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by Taran » Sat Oct 22, 2011 12:39 am
n = 8 x p

p has only two factors: 1 and p itself. Note that since p>2, it is odd and therefore does not share any common factor with 8, which is basically 2x2x2, except 1. 8 has 4 factors - 1, 2, 4 & 8.

Factors of 8 - 4
Factors of p - 2
Total - 6 + 1(8xp is also a factor)=7
1 is repeated twice, therefore total factors of 8xp is 6.

Therefore D.
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