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How many prime divisors does positive integer N have? 1) 2N

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How many prime divisors does positive integer N have? 1) 2N [#permalink] New post 16 May 2006, 07:13
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A
B
C
D
E

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How many prime divisors does positive integer N have?
1) 2N has one prime divisor
2) 3N has one prime divisor

Can you please explain.
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 [#permalink] New post 16 May 2006, 07:22
N = positive integer = {1, 2, 3, 4, ...}

Taking (1)
If 2N has one prime divisor, it has to be 2.
So N=1
Sufficient

Taking (2)
Same way, the prime divisor is 3
So, N=1
Sufficient

Hence D
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 [#permalink] New post 16 May 2006, 07:40
C it is

1) if 2N has 1 prime divisor, N=2 (1 prime divisor ) or N=1 (no prime divisors) so insuff.

2) as above so insuff.

(1) + (2) imply that N=1, so it has no prime divisors so it is sufficient
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 [#permalink] New post 16 May 2006, 07:48
B it should be.

A is insufficient as if N = 1 then 2N, i.e 2*1 = 2 has just one prime divisor.
if N = 2, then 2N, 2*2 also has ust one prime divisor i.e 2
so inufficient.

B.....3N if N = 1, then one prime divisor i.e 3
else for neother value of N, it will have more than one divisor.
hence B is sufficient.
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 [#permalink] New post 16 May 2006, 07:50
agree with deowl .... thinking too fast (meaning not thinking).
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 [#permalink] New post 16 May 2006, 07:58
deowl wrote:
C it is

1) if 2N has 1 prime divisor, N=2 (1 prime divisor ) or N=1 (no prime divisors) so insuff.

2) as above so insuff.

(1) + (2) imply that N=1, so it has no prime divisors so it is sufficient


Deowl, sorryr for being a dumbo. I still dont get it.

2N has 1 prime divisor. Doesnt that mean N can only take the value of 1?
ie. N=1 , only one prime divisor (2),
N=2, two prime divisors (2,2)
N=3, two prime divisors (2,3)

So, only N=1 satisfies this equation.

Can you pls tell me where I have gone wrong in my understanding?
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Re: DS Prime divisors [#permalink] New post 16 May 2006, 07:59
remgeo wrote:
How many prime divisors does positive integer N have?
1) 2N has one prime divisor
2) 3N has one prime divisor

Can you please explain.


i guess C as well.

from i.
if n = 2, 2n = 4 and it has only one prime divisor i.e. 2.
if n = 1, 2n = 2 and it has only one prime divisor i.e. 2.

from ii.
if n = 3, 3n = 9 and it has only one prime divisor i.e. 3.
if n = 1, 3n = 3 and it has only one prime divisor i.e. 3.

from i and ii, n should be 1.
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 [#permalink] New post 16 May 2006, 09:17
I go with B.

stmt 2: if n=2 then 3n=6 So, there are 2 primary divisors 2 and 3. therefore, n has to be 1.
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 [#permalink] New post 16 May 2006, 09:21
shampoo wrote:
I go with B.

stmt 2: if n=2 then 3n=6 So, there are 2 primary divisors 2 and 3. therefore, n has to be 1.


grrrr. but if n=3 as Prof said then 3n=9 and one prime divisor.

I am chaning my ans to C :(
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 [#permalink] New post 16 May 2006, 09:22
What would you say about the difficulty level of this question?
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 [#permalink] New post 16 May 2006, 10:50
remgeo wrote:

Deowl, sorryr for being a dumbo. I still dont get it.

2N has 1 prime divisor. Doesnt that mean N can only take the value of 1?
ie. N=1 , only one prime divisor (2),
N=2, two prime divisors (2,2)
N=3, two prime divisors (2,3)

So, only N=1 satisfies this equation.

Can you pls tell me where I have gone wrong in my understanding?



When you are counting factors (divisors) of an integer, you should only count different ones.
  [#permalink] 16 May 2006, 10:50
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How many prime divisors does positive integer N have? 1) 2N

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