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

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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.

---------------------- +1Kudos if the answer helped

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

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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.

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.
_________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

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

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

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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.

"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.
_________________

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

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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?

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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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?

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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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?