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

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Intern
Joined: 18 May 2017
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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09 Jul 2017, 07:08
1
Bunuel - i have subtracted 13! from each side of the inequality and got that k is equal or between 0 to 11. From this inequality i can't got the answer. Where am i wrong? By the way - i looked at the math book for inequality theory but haven't find something relevant. Is there any post that covers this topic?

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

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12 Nov 2017, 09:22
Hi Bunuel,

I have the same question as oryahalom.

Cant we subtract 13! from both sides of the inequality?

Thanks,
RD
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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12 Nov 2017, 09:28
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rudjlive wrote:
Hi Bunuel,

I have the same question as oryahalom.

Cant we subtract 13! from both sides of the inequality?

Thanks,
RD

Are there only two parts in $$13! + 2 \leq k \leq 13!+13$$? No, there are three. If you subtract 13! you should subtract from all three parts and you'll end up with $$2 \leq k -13! \leq 13$$, which gives you nothing.
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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19 Dec 2017, 07:46
gmatcracker2010 wrote:
Does the integer k have a factor p such that 1<p<k?

(1) k > 4!

(2) $$13! + 2 \leq k \leq 13!+13$$

We need to determine whether k has a factor p such that 1<p<k, or in other words, whether k is a prime number. If it is, then it doesn’t have a factor between 1 and itself. If it isn’t, then it does.

Statement One Alone:
k > 4!

Since there are prime numbers greater than 4! and composite (non-prime) numbers greater than 4!, statement one alone is not sufficient to answer the question.

Statement Two Alone:
13! + 2 ≤ k ≤ 13! + 13

13! will have a factor of any integers from 2 to 13 inclusive. For any number k is, in the range of integers from 13! + 2 to 13! + 13 inclusive, k will have a factor from 2 to 13 inclusive. In particular, if k = 13! + n where (2 ≤ n ≤ 13), k will have n as a factor. For example, if k = 13! + 5, then k will have a factor of 5, since 5 divides into 13! and 5. If k = 13! + 8, then k will have a factor of 8 and hence a factor of 2.

Thus, statement two is sufficient to answer the question.

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

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19 Aug 2019, 01:33
Hi Bunuel,

If we go with 2nd option then we can have multiple factors

For ex: If we take k=13!+4 then apart from 4, it will have others factors also .

So we will be getting more than one values of P.

So is it correct to mark B as sufficient answer.

Thanks
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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19 Aug 2019, 01:45
1
a12bansal wrote:
Hi Bunuel,

If we go with 2nd option then we can have multiple factors

For ex: If we take k=13!+4 then apart from 4, it will have others factors also .

So we will be getting more than one values of P.

So is it correct to mark B as sufficient answer.

Thanks

This is not a value question. The question is NOT what is the value of p.

It's an YES/NO question: Does the integer k have a factor p such that 1<p<k? And from (2) we get a definite YES answer to that question.

Hope it's clear.
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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19 Aug 2019, 02:06
gmatcracker2010 wrote:
Does the integer k have a factor p such that 1<p<k?

(1) k > 4!

(2) $$13! + 2 \leq k \leq 13!+13$$

For K >4 ! (=24)

k can be prime or composite. k = 29, 31 (which are prime) and k = 25, 26, 27 etc which are composite.
So the answer in case K is prime is No. But in case k is composite, the answer is Yes....as there will be factors which lie between the number and 1.

For 13! + 2 <= k <=13! +13, k can be anything from 13!+2, 13!+3, 13! + 4, .......until 13!+13. (in all 12 numbers).

In all these possibilities, a factor can be found which is greater than 1 and less than k.

For example. in 13! + 2, 2 can be taken as common as 13! will certainly have 2. so 2 is a factor
in 13! + 7, 7 can be taken common from 13! and 7. So 7 is a factor.
So in all these scenarios, we are sure to find a factor.

Had this been 13! + 1, it would have been difficult to find if there is any factor for this or not.

Let me know if you still have any doubt.
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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19 Aug 2019, 02:47
gmatcracker2010 wrote:
Does the integer k have a factor p such that 1<p<k?

(1) k > 4!
(2) $$13! + 2 \leq k \leq 13!+13$$

Simplifying the question, it simply asks whether integer k is non prime or not? (because in case it has no factor p such that 1<p<k then it will be prime)

Now let us see the statements

Statement (1)$$k > 4!$$ (it may or may not be prime. Hence NOT SUFFICIENT )
Strike off A and D

Statement (2) for all cases defined the number can not be a prime, as we have 2,3,4......13 respectively as a factor for each of the 12 numbers in the range defined

$$13! + 2$$ = $$2*(13*12*11*10*9*8*7*6*5*4*3*1 + 1)$$( Similarly till 13 we have a common factor of 2...13)

Hence , Clearly 2 Alone is Sufficient,

Hence B
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Re: Does the integer k have a factor p such that 1 < p < k ?   [#permalink] 19 Aug 2019, 02:47

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