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# Does the integer k have a factor p such that 1 < p <

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Intern
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Does the integer k have a factor p such that 1 < p < [#permalink]

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07 Oct 2005, 05:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Does the integer k have a factor p such that 1 < p < k?
(1) k > 4!
(2) 13! + 2 <= k <= 13! + 13

A) Statement (1) Alone is sufficient, but statment (2) alone is not sufficient.
B) Statement (2) Alone is sufficient, but statment (1) alone is not sufficient.
C) Both statements together are sufficient, but neither statement alone is sufficient.
D) Each statement alone is sufficient.
E) Statements (1) and (2) together are not sufficient.

_________________

Make use of the socratic method...

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07 Oct 2005, 06:10
The question seems to be asking if integer k is a prime number. Afterall prime numbers only have the factors of 1 and itself. So

(1) Insuff. 4! = 24 but both k = 37 and k=36 could be true so no dice.

(2) Suff. You don't need to calculate since you know that the final number has factors 2-13, adding 2 through 13 will not produce a prime number.

ie. 2 + (13*12*11*10*9*8*7*6*5*4*3*2) will still be a factor of two

and similarly 13 + (13*12*11*10*9*8*7*6*5*4*3*2) will still be a factor of 13.

Both factors have to be less than the number itself.

So I would go with (B).

Brendan.

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07 Oct 2005, 07:52
bsmith75 wrote:
The question seems to be asking if integer k is a prime number. Afterall prime numbers only have the factors of 1 and itself. So

(1) Insuff. 4! = 24 but both k = 37 and k=36 could be true so no dice.

(2) Suff. You don't need to calculate since you know that the final number has factors 2-13, adding 2 through 13 will not produce a prime number.

ie. 2 + (13*12*11*10*9*8*7*6*5*4*3*2) will still be a factor of two

and similarly 13 + (13*12*11*10*9*8*7*6*5*4*3*2) will still be a factor of 13.

Both factors have to be less than the number itself.

So I would go with (B).

Brendan.

I agree with Brendan... exactly right...

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07 Oct 2005, 09:22
Does the integer k have a factor p such that 1 < p < k?
(1) k > 4!
(2) 13! + 2 <= k <= 13! + 13

A) Statement (1) Alone is sufficient, but statment (2) alone is not sufficient.
B) Statement (2) Alone is sufficient, but statment (1) alone is not sufficient.
C) Both statements together are sufficient, but neither statement alone is sufficient.
D) Each statement alone is sufficient.
E) Statements (1) and (2) together are not sufficient.

Question can be restated as :
B.

Is K prime?

(1) : k > 4! there are infinite primes. obviously there is one > 4!. Insufficient.

(2) as we know n! is divisible by all natural number from 1 through n. Now imagin adding any number 1 to n. The new numbers will be divisible by the number added. For example:

5! = 5 * 4 * 3 * 2 * 1
5! + 2 = (5 * 4 * 2 * 1) + 2. As both parts are divisible by 2, sum is divisible by 2.
Similarly,
5! + 3 is divisible by 3

or n! + n is divisible by n

Suffucient.

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Re: Integer Problem   [#permalink] 07 Oct 2005, 09:22
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# Does the integer k have a factor p such that 1 < p <

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