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Positive integer?

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by GmatKiss » Sun Oct 16, 2011 8:49 am
How many factors does x have, if x is a positive integer?

(1) x = p^n, where p is a prime number.
(2) n^n = n + n, where n is a positive integer.

IMO:C
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by gmatclubmember » Sun Oct 16, 2011 10:44 am
GmatKiss wrote:How many factors does x have, if x is a positive integer?

(1) x = p^n, where p is a prime number.
(2) n^n = n + n, where n is a positive integer.

IMO:C
Statement 2 doesn't seems right. there is no X. can you please check it.
But Statement 1 is sufficient to answer.
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by moonraker » Sun Oct 16, 2011 10:58 am
here an additional info should have been that p is not equal to n.

If that is written then there are 4 factors including 1 and the number itself.

If not and p = n = 2 then there are 3 factors ( 2, x and 1)

Statement C may not be completely correct.[/img]
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by fcabanski » Sun Oct 16, 2011 11:03 am
1.

If x=9 p=3 n=2

3^2=9,

9 has 3 factors: 3, 9, 1

But p and n could be other values. n could be 3, in which case x would have 4 factors.

p=3 n=3 x=27 factors 27, 1, 9, 3

1 is not sufficient. It leads to multiple values for x.

2. n has to be 2, because only 2^2 = 2+2 = 4. Any other n won't fulfill that. 3^3 = 9 3+3 =6...

But there's nothing to say what relation n (2) has to x. n=2 tells nothing about x, or its factors. x could be 10, 1000, 252, 1347.

2 is not sufficient.

1 and 2.

x=p^n and n=2 indicates x must have 3 factors.

any p (prime) raised to the 2nd power will have 3 factors.

17^2=289

Factors are 289, 1, and 17

1 and 2 together are sufficient.
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by moonraker » Sun Oct 16, 2011 11:13 am
fcabanski wrote:1.

If x=9 p=3 n=2

3^2=9,

9 has 3 factors: 3, 9, 1

But p and n could be other values. n could be 3, in which case x would have 4 factors.

p=3 n=3 x=27 factors 27, 1, 9, 3

1 is not sufficient. It leads to multiple values for x.

2. n has to be 2, because only 2^2 = 2+2 = 4. Any other n won't fulfill that. 3^3 = 9 3+3 =6...

But there's nothing to say what relation n (2) has to x. n=2 tells nothing about x, or its factors. x could be 10, 1000, 252, 1347.

2 is not sufficient.

1 and 2.

x=p^n and n=2 indicates x must have 3 factors.

any p (prime) raised to the 2nd power will have 3 factors.

17^2=289

Factors are 289, 1, and 17

1 and 2 together are sufficient.


the condition p not equal to n is also a must.
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by fcabanski » Sun Oct 16, 2011 11:32 am
x=p^n

x positive integer
p prime number
n positive integer n^n=n+n n=2

p=2=n

2^2 = 4 = x

factors of x: 4, 2, 1 : 3 factors
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