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

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

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New post 13 Apr 2014, 01:10
Mountain14 wrote:
Does the integer g have a factor f such that 1 < f < g ?

1) g>3!
2) 11!+11≥g≥11!+2



What we need to figure out in this question is if g is a prime number or not.
If g is a prime number, it will not have a factor f, else it will have a factor f.

1) g > 3! means g is any number greater than 6. It can be a prime number, we do not know for sure.
2) 11!+11≥g≥11!+2 means g = 11! + k where 2 <= k <= 11, as all the numbers between 2 and 11 is present in 11!. g = k *(1+ 1*2*....)
g will always have x as factor, so it is not prime.

Hence, 2 is sufficient to answer the question. So option B.


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

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New post 13 Apr 2014, 06:01
Mountain14 wrote:
Does the integer g have a factor f such that 1 < f < g ?

1) g>3!
2) 11!+11≥g≥11!+2


Merging similar topics. Please refer to the solutions above.

Similar questions to practice:
if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.html

for-any-integer-n-greater-than-1-n-denotes-the-product-of-168575.html
does-integer-n-have-2-factors-x-y-such-that-1-x-y-n-165983.html
if-z-is-an-integer-is-z-prime-128732.html

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

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New post 06 Dec 2014, 05:06
Bunuel wrote:
M3tm4n wrote:
Hi bunuel
I don't understand the definition of a prime ( 1<p<k ). I know what a prime is but this drives me nuts.
For an example I take 10 for k. I have factors 2x5 that are greater than 1 and smaller than 10. :?:

Or is it asking for one factor? I see "a factor" in the question, which is singular. Then I understand this. It is really tricky.

Question asks whether some number k has a factor p which is more than 1 but less than k. For example if k=10 then the answer is yes, since both 2 and 5 are factors of 10 and are more than 1 and less than 10. But if for example k=7=prime then the answer is no, since 7 has no factor which is more than 1 and less than 7.

Now, look at the definition of a prime number: a prime number is a positive integer with exactly two factors: 1 and itself. So, we can say that the questions asks whether k is a prime number, because if it is then it won't have a factor which is more than 1 and less than k.

Hope it's clear.


Hi Bunuel- I thought the question asked for single value of p that satisfied the condition 1<p<k. What did I miss/ How avoid such a mistake again?
Thanks in advance.

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

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New post 06 Feb 2015, 20:25
Pretty sure there's a 3 missing in the otherwise brilliant explanation provided by Bunuel.

13!+8= 8(2*3*4*5*6*7*9*10*11*12*13+1)

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

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New post 03 Mar 2015, 09:29
enigma123 wrote:
Does the integer k have a factor p such that 1 < p < k ?

(1) k > 4!
(2) 13! + 2 <= k <= 13! + 13

[Reveal] Spoiler:
So considering statement 1

K > 24

Lets say k = 25, so the factors are 1, 5 and 25. P will be 1 as 1 is the factor of every number. Therefore, insufficient.

Considering Statement 2

13!+2<= k<=13!+13

simplifying this expression will give

2<=K<=13, so can be anything from 2...............13. As p can and cannot be greater than 1 this statement is insufficient.

Combining the two, I think will not give an answer too. Therefore, for me its E. Any thoughts anyone as I don't have an OA?


Question is asking whether K is prime ?
Option A : K>4! ; if K=25 then there exists 5 ; If K=29 ; there does not exists any integer between 1 and 29 which is factor of 29. Not Sufficient.
option B: 13!+2<= k<=13!+13 ; note here that 13! is multiple of all the numbers from 2 to 13 (inclusive) ; adding any number between 2 and 13 will just give another Even number and we know that there is only one even Prime Number 2 so clearly all integers in this range 13!+2 to 13!+13 will be Even numbers , so B is sufficient in telling that K is NOT PRIME.
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]

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New post 04 Oct 2015, 17:30
gmatcracker2010 wrote:
i dont have any idea to solve the attached problem. Please provide methodology to solve such problems.

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

(1) k > 4!
(2) 13! + 2<= k <= 13!+13

Target question: Does the integer k have a factor p such that 1 < p < k ?

This question is a great candidate for rephrasing the target question. (We have a free video with tips on rephrasing the target question: http://www.gmatprepnow.com/module/gmat-data-sufficiency?id=1100)

Let's look at a few cases to get a better idea of what the target question is asking.
- Try k = 6. Since 2 is a factor of 6, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 10 Since 5 is a factor of 10, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 16. Since 4 is a factor of 14, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 5. Since 1 and 5 are the ONLY factors of 5, we can see that k does NOT have a factor p such that 1<p<k.
Aha, so if k is a prime number, then it CANNOT satisfy the condition of having a factor p such that 1 < p < k
In other words, the target question is really asking us whether k is a non-prime integer (aka a "composite integer")

REPHRASED target question: Is integer k a non-prime integer?

Statement 1: k > 4!
In other words, k > 24
This does not help us determine whether or not k is a non-prime integer? No.
Consider these two conflicting cases:
Case a: k = 25, in which case k is a non-prime integer
Case b: k = 29, in which case k is a prime integer
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 13! + 2 ≤ k ≤ 13! + 13
Let's examine a few possible values for k.

k = 13! + 2
= (13)(12)(11)....(5)(4)(3)(2)(1) + 2
= 2[(13)(12)(11)....(5)(4)(3)(1) + 1]
Since k is a multiple of 2, k is a non-prime integer

k = 13! + 3
= (13)(12)(11)....(5)(4)(3)(2)(1) + 3
= 3[(13)(12)(11)....(5)(4)(2)(1) + 1]
Since k is a multiple of 3, k is a non-prime integer

k = 13! + 4
= (13)(12)(11)....(5)(4)(3)(2)(1) + 4
= 4[(13)(12)(11)....(5)(3)(2)(1) + 1]
Since k is a multiple of 4, k is a non-prime integer

As you can see, this pattern can be repeated all the way up to k = 13! + 13. In EVERY case, k is a non-prime integer

Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

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

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New post 28 Mar 2016, 17:21
Here is a visual that should help.
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Screen Shot 2016-03-28 at 6.20.15 PM.png [ 133.71 KiB | Viewed 1256 times ]


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

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New post 14 May 2016, 13:39
gmatcracker2010 wrote:
i dont have any idea to solve the attached problem. Please provide methodology to solve such problems.

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

(1) k > 4!
(2) 13! + 2<= k <= 13!+13


St1. we an have a prime and no prime, hence -> not suff.
St2. We need to find only 1 p that fits the discription. 13!+2=[2(13!/2+1)]
-> so 2 fits the description.
-> Suff.

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

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New post 04 Dec 2016, 20:07
"Question basically asks whether k is a prime number. If it is, then it won't have a factor psuch that 1<p<k

(definition of a prime number)."

Hi,

I just don't understand this. Can you please explain?

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

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New post 05 Dec 2016, 00:16
ruchitd wrote:
"Question basically asks whether k is a prime number. If it is, then it won't have a factor psuch that 1<p<k

(definition of a prime number)."

Hi,

I just don't understand this. Can you please explain?


A prime number is a positive integer with only two factors 1 and itself. So, a prime number does not have a factor which is more that 1 and less than itself.
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]

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New post 08 Jul 2017, 12:07
[quote="Bunuel"][quote="mneeti"][quote="Bunuel"]


Bunuel , Please make me understand one thing.

The question tells us that 1<p<k

which means p can be anything smaller than k , and question asks if p is a factor of K?

it means if k is 25 then is p a factor of 25? or we can assume p be any number less than 25 and check if its factor of 25 or not

My question is in option B,

If 13! + 2 <= k <= 13! + 13 , it means k is a very large digit and lies between those equalities.
SO NOW , p must be any integer less than k and we need to see if p is a factor of k?

How about 17? lets take p as 17

1<=17<=(13!+2,13!+3,13!+4........13!+13)
as we know from B that k has a factor , yes we can see 2, 3 ,4 ,5 till 13 it has factor but hat about 17?

17 satisfies the criteria and it should also be a factor and if its not then it must be uncertain if p is factor of k or not?


PLEASE HELP

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

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New post 09 Jul 2017, 01:16
rocko911 wrote:
Bunuel , Please make me understand one thing.

The question tells us that 1<p<k

which means p can be anything smaller than k , and question asks if p is a factor of K?

it means if k is 25 then is p a factor of 25? or we can assume p be any number less than 25 and check if its factor of 25 or not

My question is in option B,

If 13! + 2 <= k <= 13! + 13 , it means k is a very large digit and lies between those equalities.
SO NOW , p must be any integer less than k and we need to see if p is a factor of k?

How about 17? lets take p as 17

1<=17<=(13!+2,13!+3,13!+4........13!+13)
as we know from B that k has a factor , yes we can see 2, 3 ,4 ,5 till 13 it has factor but hat about 17?

17 satisfies the criteria and it should also be a factor and if its not then it must be uncertain if p is factor of k or not?


PLEASE HELP


Forget about p. The question asks does a positive integer k, has a factor which is greater than 1 and less than k itself. As explained on previous pages this is the same as asking is k a prime number. If k is a prime number it won't have any such factor but if k is not a prime number, then it'll have such factor(s).
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]

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New post 09 Jul 2017, 06:08
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?

Thanks in advance!

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

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New post 12 Nov 2017, 08:22
Hi Bunuel,

I have the same question as oryahalom.

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

Thanks,
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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,

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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New post 12 Nov 2017, 08:48
Thanks Bunuel for the immediate reply! : )

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

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New post 18 Nov 2017, 05:10
I did something wrong, which I would like to bring for everyone's attention:

When I saw this question in QP1, I hurriedly cancelled 13! from both sides, reducing the expression to:


13! + 2 ≤ k ≤ 13! + 13


Thereafter, I went on to establish that since the expression is left as 2 ≤ k ≤ 13, so we can have both primes (2, 3, 5, 7, 11, 13) and non-primes (4, 6, 8, 10, 12) in the range. So I picked in-sufficient for this statement.


Now I realize that the my method was not right. What I should have realized is that we take 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and 13 common from 2 + 13!, 3 + 13!, 4 + 13!, 5 + 13!, 6 + 13!, 7 + 13!, 8 + 13!, 9 + 13!, 10 + 13!, 11 + 13!, 12 + 13!, and 13 + 13!. And by doing so, we have proven that none of the numbers in the given range is prime.


Now I realize that my method was incorrect. Maybe my concept of crossing out terms from an inequality is flawed and I need to work on it.


Bunuel: If you don't mind, could you please let me know why we cannot cross-out from an inequality expression like I tried when attempting this question?

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

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See Bunuel's last post above

you would have to cross out 13! in all 3 parts of the inequality, while it is only present in 2

13!+2≤k≤13!+13 =

13!+2≤k
AND
k≤13!+13

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Re: Does the integer k have a factor p such that 1 < p < k ?   [#permalink] 18 Nov 2017, 09:05

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