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# Is integer k a multiple of 14?

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Re: Is integer k a multiple of 14? [#permalink]
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Bunuel wrote:
Is integer k a multiple of 14?

(1) k > 13!
(2) k = m!, where m is an integer greater than 6.

Hi,

Let's see the statements..
1) k >13!..
Don't take it to be 14!...
13! + 1 will not be div by 2 or 7, as 13! Is div by 2 and 7..
14! Will be div by 2 and 7..
Insufficient

2) k=m!, m>6 and an integer..
Next would be 7, and 7! Will contain both 2 and 7..
Anything above in the given condition will be div by 14..
Suff

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Re: Is integer k a multiple of 14? [#permalink]
answer is D. as both option include 2*7
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Re: Is integer k a multiple of 14? [#permalink]

a number > 13! is not certainly divisible by 14. so stmt-1 is insuff.

a number factorial of greater than 6 will have both 7 and 2. hence divisible by 14. suff.
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Re: Is integer k a multiple of 14? [#permalink]
(1) would be sufficient if K were an integer. But it is not mentioned.

In (2) M is an integer and M>6. So the smallest possible value of M! = K is 7! which includes (2*7). Sufficient.
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Re: Is integer k a multiple of 14? [#permalink]
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It's important to note that we DO, in fact, know that k is an integer. The question stem, "Is integer k a multiple of 14?" tells us it is.

(if the question stem were "is k a multiple of 14?" or "is k an integer that is divisible by 14?" we would not know. But with the wording the way it is, we do know.)

Nonetheless, statement one is till insufficient. All we are told is that (integer) k is greater than some large number. But only every fourteenth integer will be on the "14 times table". In other words some of the integers greater than 13! will be a multiple of 14 and some of them won't. 14!, for example will be a multiple of 14, but the next integer up, 14! + 1, will not. Sometimes Yes, sometimes No; Statement 1 is insufficient.

As others have pointed out, the best way to see why statement two is sufficient is to consider prime factorization. If k/14 is going to be an integer, then k/(2*7) will be an integer. And if k/(2*7) is to be an integer, then the denominator must cancel out completely. In other words, k must include a 2 and a 7 in its prime factorization. Statement 2 tells us k could be 7! or 8! or 9! (and so on). No matter what, k will have at least one 2 and at least one 7 in its PF. Sufficient.

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Re: Is integer k a multiple of 14? [#permalink]
Given : k is an integer.

1. k>13!

13! is divisible by 14.(13x12x11x10x9x8x7x....)
But k is greater than 13!
So, possibilities :

k = 13! + 1(Not divisible by 14)
or
k = 13! + 14( Divisible by 14)

2. k = m!, where m is an integer greater than 6.

k has a definite value that is equal to m!
And m > 6(m=7, 8,9,10)
Possibilities:

m = 7:

k = 7! or 8! or 9! (Divisible by 14, as 7! = 7x6....).

SUFFICIENT