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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.

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

Question basically asks is k 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). Sufficient.

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

Question basically asks is k 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). Sufficient.

Thanks Bunuel. I have a question, what would happen if the second statement said this?

(2) 13!+1\leq{k}\leq{13!+13}

I have seen some similar problems, in which they add 1. Could you post some links about it?

Thanks!

So basically we should determine whether 13!+1 is a prime number (as shown above all other possible values of k are not are not prime), which can not be done without a computer. There are some particular values of k=n!+1 for which we can say whether it's a prime or not with help of Wilson's theorem, but again it's out of the scope of GMAT.

By the way: 13!+1 is not a prime number, it has two distinct prime factors: 13!+1=83*75,024,347=6,227,020,801, so the answer still will be B.

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). Sufficient.

Sorry for the question, but I am just not understanding how you get 8*(2*4*5*6*7*9*10*11*12*13+1) from 13!+8 then how do you know that k is a multiple of 8...I obviously have some deficiencies when it comes to number properties. So then how is k never a prime between those parameters? I would think (1*2*3*4*5*6*7*8*9*10*11*12*13) +8 ...

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). Sufficient.

Sorry for the question, but I am just not understanding how you get 8*(2*4*5*6*7*9*10*11*12*13+1) from 13!+8 then how do you know that k is a multiple of 8...I obviously have some deficiencies when it comes to number properties. So then how is k never a prime between those parameters? I would think (1*2*3*4*5*6*7*8*9*10*11*12*13) +8 ...

k=13!+8 means k=2*3*4*5*6*7*8*9*10*11*12*13+8, now factor out 8: k=8*(2*4*5*6*7*9*10*11*12*13+1) --> k is a multiple of 8 as k=8*something --> as k is a multiple of 8 it can not be a prime number. You can have any number from 2 to 13 inclusive instead of 8, and you'll be able to factor out this number the same way as you did with 8, so any number of a type 13!+x, wher x is from 2 to 13 inclusive will be a multiple of x, thus not a prime number.

Check the link in my first post for similar problem. Also check Number Theory chapter of Math Book: math-number-theory-88376.html

Answer can't be E, it has to be B. I don't have OG with me, but I need to check.

1. By itself not sufficient. K could be prime. 2. 13! + 2 <= K <= 13! + 13 ensures that you have ever number from 2 - 13 dividing k. That is: 13! + 2 -> div by 2 13! + 3 -> div by 3 ... so on till 13.

So, answer should be B.

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I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min