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Re: K have a factor p [#permalink]
29 Jan 2012, 15:21

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

Question basically asks whether k is a prime number. If it is, then it won't have a factor p such that 1<p<k (definition of a prime number).

(1) k>4! --> k is more than some number (4!=24). k may or may not be a prime. Not sufficient.

(2) 13!+2\leq{k}\leq{13!+13} --> k can not be a prime. For instance if k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1), then k is a multiple of 8, so not a prime. Same for all other numbers in this range. So, k=13!+x, where 2\leq{x}\leq{13} will definitely be a multiple of x (as we would be able to factor out x out of 13!+x, the same way as we did for 8). Sufficient.

Re: Does the integer k have a factor p such that 1<p<k? [#permalink]
14 Feb 2012, 13:27

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.

Re: Does the integer k have a factor p such that 1<p<k? [#permalink]
14 Feb 2012, 15:15

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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.

Re: K have a factor p [#permalink]
18 Nov 2012, 02:18

Bunuel wrote:

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

Question basically asks whether k is a prime number. If it is, then it won't have a factor p such that 1<p<k (definition of a prime number).

(1) k>4! --> k is more than some number (4!=24). k may or may not be a prime. Not sufficient.

(2) 13!+2\leq{k}\leq{13!+13} --> k can not be a prime. For instance if k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1), then k is a multiple of 8, so not a prime. Same for all other numbers in this range. So, k=13!+x, where 2\leq{x}\leq{13} will definitely be a multiple of x (as we would be able to factor out x out of 13!+x, the same way as we did for 8). Sufficient.

Hi Bunuel, I have a small query here- Had the question been : 4!+2<k<4!+6, if I follow the approach explained above I may conclude that k is not a prime no. reason being all the factors of 4! are the factors of 4!+?. Since its easy to calculate here the values between 4!+2 and 4!+6, I already know that it includes 29 which is a prime no., where as in case of 13!+2<k<13!+13, it is not easy to calculate 13!, we may be missing some value which is a prime no. How do we make sure that we are not missing anything? (Assuming we can not cross check by calculating the values because of the time constraint at the exam time)

Re: K have a factor p [#permalink]
18 Nov 2012, 04:01

Expert's post

mneeti wrote:

Bunuel wrote:

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

Question basically asks whether k is a prime number. If it is, then it won't have a factor p such that 1<p<k (definition of a prime number).

(1) k>4! --> k is more than some number (4!=24). k may or may not be a prime. Not sufficient.

(2) 13!+2\leq{k}\leq{13!+13} --> k can not be a prime. For instance if k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1), then k is a multiple of 8, so not a prime. Same for all other numbers in this range. So, k=13!+x, where 2\leq{x}\leq{13} will definitely be a multiple of x (as we would be able to factor out x out of 13!+x, the same way as we did for 8). Sufficient.

Hi Bunuel, I have a small query here- Had the question been : 4!+2<k<4!+6, if I follow the approach explained above I may conclude that k is not a prime no. reason being all the factors of 4! are the factors of 4!+?. Since its easy to calculate here the values between 4!+2 and 4!+6, I already know that it includes 29 which is a prime no., where as in case of 13!+2<k<13!+13, it is not easy to calculate 13!, we may be missing some value which is a prime no. How do we make sure that we are not missing anything? (Assuming we can not cross check by calculating the values because of the time constraint at the exam time)

If 13!+2\leq{k}\leq{13!+13}, then we can factor out x out of 13!+x (where 2\leq{x}\leq{13}), which means that k is not a prime.

But you cannot apply the same logic with 4!+2<k<4!+6, since if k=4!+5, then you cannot factor out 5 out of it. _________________

Re: Does the integer k have a factor p such that 1<p<k? [#permalink]
27 Jan 2013, 20:25

So basically in the second statement we can factor out a number from 2-13 but if its higher than that we can't say for sure it would be a prime or not? _________________

The question is basically asking whether K is prime or not. If it is prime then P can be 1 or k else it will follow 1<p<k.

Statement 1. k>4! k>24; let's say k=25, then p=5; we get a Yes to the question. if k=29; then p=1 or 29; we get a No to the question. Thus, insufficient.

Statement 2: 13!+2<=k<=13!+13 k can take any value from 13!+2 to 13!+13. All the values in this range has one thing in common. They all can have one common value: for e.g. 13!+2= 2[(13*12*...except 2)+1] 13!+10=10[(13*12*...(except 10)+1] No matter whether the value in the bracket is prime or not; the common value will make it non prime and will always give a yes to the question. Thus, sufficient.

Essentially this is about prime and non-prime. Option 1: k>4! Now assume k=5. then is there any factor of 5 which lies between 1 and 5? NO. Take k=6. then there is 2,3 which lies between 1 and 6. So 1 alone insufficient.

Option 2: 13!+2<=k<=13!+13 Take say 13!+3. now it can be rewritten as 3*{(13.12.11.....4.2.1)+1} So there is 3 as factor between 1 and k. Similarly you can take the common factor out of all nos. in option 2. Therefore option 2 is sufficient! Add Kudos if this helps!

Re: Does the integer g have a factor f [#permalink]
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