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Is integer k a multiple of 14?
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20 Jan 2017, 07:25
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51% (01:06) correct 49% (00:47) wrong based on 105 sessions
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Is integer k a multiple of 14? (1) k > 13! (2) k = m!, where m is an integer greater than 6.
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Is integer k a multiple of 14?
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20 Jan 2017, 07:29
1) Minimum value of k = 14 and also 14! contains a 2 and 7 as its factors. YES. Sufficient. 2) Min value of k = 7 and 7! contains a 2 and 7 as its factors. YES. Sufficient.
D.



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Re: Is integer k a multiple of 14?
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21 Jan 2017, 05:00
Is k a multiple of 14?
Rephrase the question to be : K/14 = Integer
(1) k > 13!
K is any number that is GREATER than 13! so it does not mean to be integer.
K =14!.......Answer to question is Yes
K= 13!+ 0.5. it means that we have Number xyz.5.........So answer to question is NO
Insufficient
(2) k = m!, where m is an integer greater than 6.
As m is integer, k will always be integer.
m>6...i.e. 7, 8, 9
So we have always at least 2*7..............Answer to question is always Yes
Sufficient
Answer: B



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Re: Is integer k a multiple of 14?
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21 Jan 2017, 05:54
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.. Different answers.. 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 B
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Re: Is integer k a multiple of 14?
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21 Jan 2017, 08:02
answer is D. as both option include 2*7



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Re: Is integer k a multiple of 14?
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22 Jan 2017, 10:17
Answer should be B. a number > 13! is not certainly divisible by 14. so stmt1 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?
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22 Jan 2017, 10:45
(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?
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22 Jan 2017, 11:20
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.
B.



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Re: Is integer k a multiple of 14?
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18 Mar 2018, 00:44
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)
So by contradiction, INSUFFICIENT
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
Answer  B




Re: Is integer k a multiple of 14?
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