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If N is a positive integer, not including N, how many

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If N is a positive integer, not including N, how many [#permalink]

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If N is a positive integer, not including N, how many factors does N have?

(1) Not including N^2, N^2 has 4 factors.

(2) Not including 2N, 2N has 3 factors.
[Reveal] Spoiler: OA

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Re: Number Property DS [#permalink]

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New post 27 Oct 2008, 03:28
Is it D?

stmt1: Excluding N^2, it has 4 factors. Two of these will be 1 and N. Two other factors are possible only when N is the square of a prime number. Suppose, N = p^2, then two other factors of N^2 will be p and N*p. Hence, sufficient.

stmt2: Excluding 2N, N has 3 factors. These will be 1, 2, N. That means, N itself is a prime number. Hence, sufficient.
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Re: DS - number theory [#permalink]

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New post 03 Nov 2008, 02:41
B.

From stmt1: N^2 has 1, N and two more factors (excluding N^2). Both these factors could also be of N or, only one of them is of N. Hence, insufficient.

From stmt2: 2N has 1, 2 and N as factors (excluding 2N). Hence, N must be a prime number. Hence, sufficient.
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Re: DS - number theory [#permalink]

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New post 03 Nov 2008, 03:53
scthakur wrote:
B.

From stmt1: N^2 has 1, N and two more factors (excluding N^2). Both these factors could also be of N or, only one of them is of N. Hence, insufficient.

From stmt2: 2N has 1, 2 and N as factors (excluding 2N). Hence, N must be a prime number. Hence, sufficient.


My friend in my opinion stmnt 1 is SUFF.....and 2 is also SUFF...so i picked D but OA is A
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Re: DS - number theory [#permalink]

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New post 03 Nov 2008, 08:18
GODSPEED wrote:
If N is a positive integer, not including N, how many factors does N have?

A - Not including N^2, N^2 has 4 factors
B - Not including 2N, 2N has 3 factors



so lets assume N=a^x b^y where a and b are prime factors so no of factors is (x+1)(y+1)

1)
anyways N^2 has 4 factors not inlcuding N^2 means N has only one prime factor

lets see, N^2=16, then factors are 1, 2, 4, 8 And 16 <--N^2

if N^2=81, then factors are 1, 3, 9, 27...

N has 3 factors given..

2) 2N has factors i.e 2N i.e. 2^1 so N has to be 3 or 5..SUff

I will go with D.
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Re: DS - number theory [#permalink]

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New post 03 Nov 2008, 10:55
GODSPEED wrote:
scthakur wrote:
B.

From stmt1: N^2 has 1, N and two more factors (excluding N^2). Both these factors could also be of N or, only one of them is of N. Hence, insufficient.

From stmt2: 2N has 1, 2 and N as factors (excluding 2N). Hence, N must be a prime number. Hence, sufficient.


My friend in my opinion stmnt 1 is SUFF.....and 2 is also SUFF...so i picked D but OA is A


I need a break :oops: making too many silly mistakes today.

Yes, OA should be A.

In stmt2: if N = 4, 2N =8 and it has 1, 2, 4 as factors (excluding 8). But, N has 1,2,4 as factors.
However, if N = 3, 2N = 6 and it has 1, 2,3 as factors (excluding 6). However, N has only 1 and 3 as factors. Hence, insufficient.
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Re: DS - number theory [#permalink]

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New post 03 Nov 2008, 11:05
stmnt 2..if 2N=6 then say N=3, factors of 6 are 1, 2, 3 6<--not included , N=3 factors are 1 and 3..

if 2N=8 then factors are 1,2,4 8<--not included N=2 factors are 1 and 2

all stmnt 2 is say N is a prime number

I think D.

scthakur wrote:
GODSPEED wrote:
scthakur wrote:
B.

From stmt1: N^2 has 1, N and two more factors (excluding N^2). Both these factors could also be of N or, only one of them is of N. Hence, insufficient.

From stmt2: 2N has 1, 2 and N as factors (excluding 2N). Hence, N must be a prime number. Hence, sufficient.


My friend in my opinion stmnt 1 is SUFF.....and 2 is also SUFF...so i picked D but OA is A


I need a break :oops: making too many silly mistakes today.

Yes, OA should be A.

In stmt2: if N = 4, 2N =8 and it has 1, 2, 4 as factors (excluding 8). But, N has 1,2,4 as factors.
However, if N = 3, 2N = 6 and it has 1, 2,3 as factors (excluding 6). However, N has only 1 and 3 as factors. Hence, insufficient.
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Re: DS - number theory [#permalink]

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New post 03 Nov 2008, 11:14
fresinha12 wrote:
stmnt 2..if 2N=6 then say N=3, factors of 6 are 1, 2, 3 6<--not included , N=3 factors are 1 and 3..

if 2N=8 then factors are 1,2,4 8<--not included N=2 factors are 1 and 2
all stmnt 2 is say N is a prime number

I think D.


Here, N will be 4 and its factors will be 1,2,4.
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Re: DS - number theory [#permalink]

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New post 03 Nov 2008, 11:25
looks like you are not the only one having a bad day...

agree with A as ans..
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Re: If N is a positive integer, not including N, how many [#permalink]

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New post 07 Feb 2014, 18:17
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If N is a positive integer, not including N, how many factors does N have?
(1) Not including N^2, N^2 has 4 factors -> N^2 (a perfect square) has a total of 5 factors => N has 3 factors in total (N is a perfect square of prime number).
example: N = 4; N^2 = 16 (5 factors); N has 2 factors excluding N.
example: N = 9; N^2 = 81 (5 factors); N has 2 factors excluding N.
(2) Not including 2N, 2N has 3 factors -> 2N has a total of 4 factors
For 2N to have a total of 4 factors, N can be 2^2 or (any other prime)^1
example: N = 4; 2N = 8 (4 factors);N has 2 factors excluding N.
example: N = 3; 2N = 6 (4 factors); N has 1 factor excluding N.
NOT SUFFICIENT.
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Re: If N is a positive integer, not including N, how many [#permalink]

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Re: If N is a positive integer, not including N, how many [#permalink]

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New post 24 Aug 2016, 03:12
Tremendous Question
here we ned to get the factors of N (forget the wording except N itself as if we have the factors of N we can decrease it by one to get this value)
Statement 1 => N^2 has 4+1=>5 Factors
Also if N is a positive integer => N^2= perfect square
5=1*5
so it must have only one prime
And as N and N^p have the same prime factors => N must be prime itself => 2 factors => Suff
Statement 2 => Here if N is odd=> N will have exactly half the factors as 2N
But wait wait if N=4 => 2N still has 4 factors in here so N=4
And if N=5=> 2N=10 => 4 factors => N=5
Contradictory
Not suff
Smash that A
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Re: If N is a positive integer, not including N, how many   [#permalink] 24 Aug 2016, 03:12
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