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# If the integer n is greater than 1, is n equal to 2?

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Intern
Joined: 12 Feb 2009
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If the integer n is greater than 1, is n equal to 2? [#permalink]

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13 May 2009, 08:53
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I don't why this isn't making sense to me.

If the integer n is greater than 1, is n equal to 2?

1) n has exactly two positive factors.
2) The difference of any two distinct positive factors of n is odd.

1 just says n is prime so obv NS, but can't 2 be any even number? ie 4-1=3? I think I'm just misunderstanding the question.
Senior Manager
Joined: 08 Jan 2009
Posts: 317
Re: DS OG #132 [#permalink]

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13 May 2009, 11:56
Stmt 1 :
n can be any prime number.
Not sufficient.
Stmt 2 :
diff b/w any two distinct factors of n is odd.
n has to be 2.
if n = 4 factors ( 1,2,4)
u can see some diff will be even.
So B.
Intern
Joined: 12 Feb 2009
Posts: 24
Re: DS OG #132 [#permalink]

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13 May 2009, 16:07
Oh I see ANY two factors (ie ALL factors of n). Ok that makes sense. Just missed that.
Manager
Joined: 14 May 2009
Posts: 189
Schools: Stanford, Harvard, Berkeley, INSEAD
Re: DS OG #132 [#permalink]

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14 May 2009, 01:09
If the integer n is greater than 1, is n equal to 2?

1) n has exactly two positive factors.
2) The difference of any two distinct positive factors of n is odd.

Formulate:
N>1, Question:(N==2?)

(1) N has exactly two positive factors
<==> N is prime.

Insufficient, as N=2,3,5,7,11,13,etc.

(2) Let's find the pattern.

N=2: Factors are 1&2. Yes, the difference is odd. N=2.

N=3: Factors are 1&3: No, difference is even. Doesn't work, N != 3.

N=4: Factors are 1/2/4, difference between 2&4 is 2, even-- doesn't work, N != 4.

N=5: Factors are 1/5, difference is 4, again even-- doesn't work, N != 5.

N=6: Factors are 1/2/3/6, difference between 2/6 is even, doesn't work, N != 6.

etc...

I think we've found the pattern... N can only be 2. Sufficient.

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Director
Joined: 23 May 2008
Posts: 760
Re: DS OG #132 [#permalink]

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06 Jun 2009, 15:57
for statement 2,

suppose n=10 factors include 1,2,5,10

5-2=3, which is odd, making B insufficient.

am I missing something?
Manager
Joined: 14 May 2009
Posts: 189
Schools: Stanford, Harvard, Berkeley, INSEAD
Re: DS OG #132 [#permalink]

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06 Jun 2009, 16:08
but diff b/w 10 & 2 is 8 which is even so 10 does't work
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Intern
Joined: 28 Mar 2009
Posts: 23
Re: DS OG #132 [#permalink]

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06 Jun 2009, 16:34
Consider statement 2 by itself:

If n is any odd number then the difference between itself and 1 is always even. So n can't be any odd number.

If n is any even number then the difference between itself and 2 (since n is even, 2 is a factor) is even. So n can't be any even number.

The condition 'distinct' leaves the only possibility of 2.
Director
Joined: 23 May 2008
Posts: 760
Re: DS OG #132 [#permalink]

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06 Jun 2009, 16:50
2) The difference of any two distinct positive factors of n is odd.

arent 5 and 2 factors of 10?
whats the definition of a "distinct" factor?
Manager
Joined: 14 May 2009
Posts: 189
Schools: Stanford, Harvard, Berkeley, INSEAD
Re: DS OG #132 [#permalink]

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06 Jun 2009, 17:06
It said any

if we take 10 & 2 they difference is even
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Director
Joined: 23 May 2008
Posts: 760
Re: DS OG #132 [#permalink]

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06 Jun 2009, 22:33
It said any

if we take 10 & 2 they difference is even

ok..i miss interpreted "any" thanks!
Manager
Joined: 11 Aug 2008
Posts: 137
Re: DS OG #132 [#permalink]

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08 Jun 2009, 19:27
1) n can be any prime number => not suf
2) if n=prime or odd, the odd divisor of n subtract 1 will be even
If n=even, 2 must be a divisor of n, so n subtract 2 will be even
So the only solution is n=2 satisfy
Hence B
Re: DS OG #132   [#permalink] 08 Jun 2009, 19:27
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# If the integer n is greater than 1, is n equal to 2?

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