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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Problem Solving - Numbers properties
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anshumishra
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n = 2^3*pkoby_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
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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Anshu
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Anshu
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anshumishra
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Another waykoby_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^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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Anshu
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towerSpider
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anshumishra
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Sure.towerSpider wrote:Anshu, i didnt get your method. Can you explain?
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.
Thanks
Anshu
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Anshu
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GMATGuruNY
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We can plug in a value for p.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
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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nipunkathuria
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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
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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anshumishra
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Right, The part in bold is also the number of odd factors, as I have already posted in a post above :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
No. of even factors= All factors - odd factors= (3+1)*(1+1) - (1+1) = 8 - 2 = 6.
Thanks
Anshu
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Anshu
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thebigkats
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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
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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Everest
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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.
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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sushantgupta
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gmatblood
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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!!
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!!
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?
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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artistocrat
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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!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?
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Taran
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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.
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.
