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Math Expert V
Joined: 02 Sep 2009
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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2
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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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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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
Math Expert V
Joined: 02 Sep 2009
Posts: 64947
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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

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the question is asking whether k has a factor that is greater than 1, but less than itself.
if you're good at these number property rephrasings, then you can realize that this question is equivalent to "is k non-prime?", which, in turn, because it's a data sufficiency problem (and therefore we don't care whether the answer is "yes" or "no", as long as there's an answer), is equivalent to "is k prime?".
but let's stick to the first question - "does k have a factor that's between 1 and k itself?" - because that's easier to interpret, and, ironically, is easier to think about (on this particular problem) than the prime issue.

--

key realization:
every one of the numbers 2, 3, 4, 5, ..., 12, 13 is a factor of 13!.

this should be clear when you think about the definition of a factorial: it's just the product of all the integers from 1 through 13. because all of those numbers are in the product, they're all factors (some of them several times over).

--

consider the lowest number allowed by statement 2: 13! + 2.
note that 2 goes into 13! (as shown above), and 2 also goes into 2. therefore, 2 is a factor of this sum (answer to question prompt = "yes").

consider the next number allowed by statement 2: 13! + 3.
note that 3 goes into 13! (as shown above), and 3 also goes into 3. therefore, 3 is a factor of this sum (answer to question prompt = "yes").

etc.
all the way to 13! + 13.
works the same way each time.
so the answer is "yes" every time --> sufficient.

--

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

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Bunuel - would it be correct to say that all factorials are composite numbers?
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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anindhya25 wrote:
Bunuel - would it be correct to say that all factorials are composite numbers?

Yes, but except 0! = 1, 1! = 1, and 2! = 2.
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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Hi Bunuel - can you confirm whether or not it is incorrect to simply substitute x for y? 2 - 3/2 = -1/2
Math Expert V
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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LaurenGol wrote:
Hi Bunuel - can you confirm whether or not it is incorrect to simply substitute x for y? 2 - 3/2 = -1/2

I think you are asking about some other question. Also, 2 - 3/2 = 1/2, not -1/2.
_________________ Re: Does the integer k have a factor p such that 1 < p < k ?   [#permalink] 07 Jun 2020, 23:29

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