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Factors prob

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hariharakarthi
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Factors prob [#permalink] New post   15 Aug 2009, 07:25
00:00
A
B
C
D
E

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 2 sessions
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if n is an integer, then n is divisible by how many positive integer?
1. n =(2^x) +1, where x is an integer
2. x<5



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Re: Factors prob [#permalink] New post   Updated on: 15 Aug 2009, 10:16
SILLY MISTAKES ARE HUNTING ME :evil:

Originally posted by yezz on 15 Aug 2009, 07:51.
Last edited by yezz on 15 Aug 2009, 10:16, edited 2 times in total.
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Re: Factors prob [#permalink] New post   15 Aug 2009, 08:11
if n is an integer, then n is divisible by how many positive integer?

1. n =(2^x) +1, where x is an integer

If x = 3, (2^x)+1 = 9 = n

9 is divisible by 1, 3 and 9

If x = 4, (2^x)+1 = 17 = n

17 is prime, divisible only by itself and 1

Since you get two different results (1) is INSUFFICIENT


2. x<5

x<5 doesn't tell us anything about n
INSUFFICIENT

(1)&(2) Together.
We already did two values of x<5 in (1) so therefore (1)&(2) together is also INSUFFICIENT

ANSWER: E
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Re: Factors prob [#permalink] New post   Updated on: 15 Aug 2009, 10:15
MY BAD :oops:

Originally posted by yezz on 15 Aug 2009, 08:54.
Last edited by yezz on 15 Aug 2009, 10:15, edited 1 time in total.
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Re: Factors prob [#permalink] New post   15 Aug 2009, 09:06
I don't understand why its not E. If you plug in X is any number but 3 yes you only get 2 positive divisors, but if you plus in 3, you get 3 positive divisors9,3,1

Am I missing something further?
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Re: Factors prob [#permalink] New post   15 Aug 2009, 09:08
yezz wrote:
shanewyatt wrote:
if n is an integer, then n is divisible by how many positive integer?

1. n =(2^x) +1, where x is an integer

If x = 3, (2^x)+1 = 9 = n

9 is divisible by 1, 3 and 9

If x = 4, (2^x)+1 = 17 = n

17 is prime, divisible only by itself and 1

Since you get two different results (1) is INSUFFICIENT


2. x<5

x<5 doesn't tell us anything about n
INSUFFICIENT

(1)&(2) Together.
We already did two values of x<5 in (1) so therefore (1)&(2) together is also INSUFFICIENT

ANSWER: E


Plz re read the question .. it is a classic trick... the question is not asking to identify what n is ...the question is simply asking for the number of +ve divisors


Please correct me if I am missing something.

If x = 3, (2^x)+1 = 9 = n

9 is divisible by 1, 3 and 9

If x = 4, (2^x)+1 = 17 = n

17 is prime, divisible only by itself and 1

For x = 3 which is <5 there are 3 positive divisors: 1, 3, and 9.
For x = 4 which is <5 there are 2 positive divisors: 1, and 17.

I think you are right for all values of x<5 not including 3. Is there a reason you did not include 3?
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Re: Factors prob [#permalink] New post   15 Aug 2009, 10:14
shanewyatt wrote:
yezz wrote:
shanewyatt wrote:
if n is an integer, then n is divisible by how many positive integer?

1. n =(2^x) +1, where x is an integer

If x = 3, (2^x)+1 = 9 = n

9 is divisible by 1, 3 and 9

If x = 4, (2^x)+1 = 17 = n

17 is prime, divisible only by itself and 1

Since you get two different results (1) is INSUFFICIENT


2. x<5

x<5 doesn't tell us anything about n
INSUFFICIENT

(1)&(2) Together.
We already did two values of x<5 in (1) so therefore (1)&(2) together is also INSUFFICIENT

ANSWER: E


Plz re read the question .. it is a classic trick... the question is not asking to identify what n is ...the question is simply asking for the number of +ve divisors


Please correct me if I am missing something.

If x = 3, (2^x)+1 = 9 = n

9 is divisible by 1, 3 and 9

If x = 4, (2^x)+1 = 17 = n

17 is prime, divisible only by itself and 1

For x = 3 which is <5 there are 3 positive divisors: 1, 3, and 9.
For x = 4 which is <5 there are 2 positive divisors: 1, and 17.

I think you are right for all values of x<5 not including 3. Is there a reason you did not include 3?



THE CLASSIC MISTAKE IS MINE U R RIGHT , APOLOGIES
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Re: Factors prob [#permalink] New post   13 Sep 2009, 19:04
Did we earn the OA yet?
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Re: Factors prob [#permalink] New post   14 Sep 2009, 10:37
kylexy wrote:
mendelay wrote:
Did we earn the OA yet?

it's E

like said before its a tricky one...the best method is to read the question each time you with single option..
using AD/BCE split...

remove AD after first option....remove B after the second...

ur left with either C or E ...if its C u will not be able to find the no of factors....

so its E...

+ kudos if u like my answer!!! :)



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Re: Factors prob [#permalink]
14 Sep 2009, 10:37
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