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Manager  Joined: 16 Feb 2010
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If x is an integer, does x have a factor n such that 1 <  [#permalink]

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If x is an integer, does x have a factor n such that 1 < n < x?

(1) x > 3!

(2) 15! + 2 ≤ x ≤ 15! + 15

I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............
Veritas Prep Representative G
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I love this question, so I'll chime in even if only to bump it to the top so more people see it!

Hopefully most can see pretty quickly that x > 3! just means x > 6, and that isn't nearly enough to tell us whether it is prime.

Statement 2 is pretty neat, though: 15! + 2 ≤ x ≤ 15! + 15

Think about 15!. 15! is 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1, which means that:

15! is a multiple of 2
15! is a multiple of 3
15! is a multiple of 4
etc....
15! is a multiple of 15

Now think about multiples of 2. Every SECOND number is a multiple of 2: 2, 4, 6, 8, 10. You create multiples of 2 by adding 2 to a previous multiple of 2. If a number is even, adding 2 "keeps it" even.

The same holds for 3. Every THIRD number (3, 6, 9, 12, 15, 18) is a multiple of 3. If you have a multiple of 3 and add 3 to it, it's still a multiple of 3.

This will hold for all of these numbers - because 15! is a multiple of every number between 2 and 15, then adding any number in that range of 2-15 will ensure that the new number remains a multiple of that number, and the new number will not be prime.

Therefore, statement 2 is sufficient, and the correct answer is B.

Now...consider the number 15! + 1. It's too big a number to know offhand whether it's prime, but we do know that it is NOT a multiple of any numbers 2-15. In order to be a multiple of 2, we'd have to add a multiple of 2 to 15!, and 1 breaks us off of that every-second-number cycle. Same for 3 - we'd need to add a multiple of 3 in order to keep 15! on that every-third cycle, so 15! is not a multiple of 3. We can prove that 15! + 1 is not divisible by any numbers between 2 and 15. It's largest prime factor must then be 17 or greater.

Understanding that ideology and being able to determine divisibility of large numbers can be quite helpful on questions that might otherwise seem impossible. Thanks for posting this question!
_________________
Brian

Curriculum Developer, Instructor, and Host of Veritas Prep On Demand

Save $100 on live Veritas Prep GMAT Courses and Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews ##### General Discussion Math Expert V Joined: 02 Sep 2009 Posts: 64317 If x is an integer, does x have a factor n such that 1 < [#permalink] ### Show Tags 20 19 zisis wrote: If x is an integer, does x have a factor n such that 1 < n < x? (1) x > 3! (2) 15! + 2 ≤ x ≤ 15! + 15 I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............ If x is an integer, does x have a factor n such that 1 < n < x? Question basically asks: is $$x$$ a prime number? If it is, then it won't have a factor $$n$$ such that $$1<n<x$$ (definition of a prime number). (1) $$x>3!$$ --> $$x$$ is more than some number (3!). $$x$$ may or may not be a prime. Not sufficient. (2) $$15!+2\leq{x}\leq{15!+15}$$ --> $$x$$ cannot be a prime. For instance if $$x=15!+8=8*(2*3*4*5*6*7*9*10*11*12*13*14*15+1)$$, then $$x$$ is a multiple of 8, so not a prime. Same for all other numbers in this range: $$x=15!+k$$, where $$2\leq{k}\leq{15}$$ will definitely be a multiple of $$k$$ (as weould be able to factor out $$k$$ out of $$15!+k$$). Sufficient. Answer: B. _________________ Manager  Joined: 17 Nov 2009 Posts: 212 Re: factor factorials [#permalink] ### Show Tags zisis wrote: If x is an integer, does x have a factor n such that 1 < n < x? (1) x > 3! (2) 15! + 2 ≤ x ≤ 15! + 15 I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............ Rephrase: Is there a factor of x that is greater than 1? So x should not be prime 1. if x> 3! then x>6 but 7 is a prime and greater than 6 and 9 is not prine but greater than 6. No solid answer 2. If x lies between 15!+2 and 15!+ 15 then there are no primes here.. B Manager  Joined: 20 Jul 2010 Posts: 166 Re: factor factorials [#permalink] ### Show Tags Nice rephrase. Somehow in option 1 I took my values as 3!, 4!, 5! and thought they all will have a factor as needed. Manager  Joined: 16 Feb 2010 Posts: 154 Re: factor factorials [#permalink] ### Show Tags Bunuel wrote: zisis wrote: If x is an integer, does x have a factor n such that 1 < n < x? (1) x > 3! (2) 15! + 2 ≤ x ≤ 15! + 15 I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............ If x is an integer, does x have a factor n such that 1 < n < x? Question basically asks: is $$x$$ a prime number? If it is, then it won't have a factor $$n$$ such that $$1<n<x$$ (definition of a prime number). (1) $$x>3!$$ --> $$x$$ is more than some number (3!). $$x$$ may or may not be a prime. Not sufficient. (2) $$15!+2\leq{x}\leq{15!+15}$$ --> $$x$$ can not be a prime. For instance if $$x=15!+8=8*(2*3*4*5*6*7*9*10*11*12*13*14*15+1)$$, then $$x$$ is a multiple of 8, so not a prime. Same for all other numbers in this range: $$x=15!+k$$, where $$2\leq{k}\leq{15}$$ will definitely be a multiple of $$k$$ (as weould be able to factor out $$k$$ out of $$15!+k$$). Sufficient. Answer: B. I am able to follow until the point that I have highlighted red....... tried to work your point my self so decided to find if x for 5!+3<x<5!+8 works with your statement above.... so - $$5! = 120$$ $$5!+3 / 3 = 123 / 3 =41$$correct $$5!+4 / 4 = 124 / 4 =31$$ correct $$5!+5 / 5 = 125 / 5 =25$$ correct $$5!+6 / 6 = 126 / 6 =21$$correct $$5!+7 / 7 = 127 / 7 =18.14$$INCORRECT !!!! so...? what am i doing wrong??? Math Expert V Joined: 02 Sep 2009 Posts: 64317 If x is an integer, does x have a factor n such that 1 < [#permalink] ### Show Tags 1 zisis wrote: I am able to follow until the point that I have highlighted red....... tried to work your point my self so decided to find if x for 5!+3<x<5!+8 works with your statement above.... so - $$5! = 120$$ $$5!+3 / 3 = 123 / 3 =41$$correct $$5!+4 / 4 = 124 / 4 =31$$ correct $$5!+5 / 5 = 125 / 5 =25$$ correct $$5!+6 / 6 = 126 / 6 =21$$correct $$5!+7 / 7 = 127 / 7 =18.14$$INCORRECT !!!! so...? what am i doing wrong??? You replaced 15! by 5!. Thus you cannot factor out 7 out of 5!+8 and this is the whole point here. If you want to check with smaller numbers try $$5!+2\leq{x}\leq{5!+5}$$ _________________ Manager  Joined: 16 Feb 2010 Posts: 154 Re: factor factorials [#permalink] ### Show Tags Bunuel wrote: zisis wrote: I am able to follow until the point that I have highlighted red....... tried to work your point my self so decided to find if x for 5!+3<x<5!+8 works with your statement above.... so - $$5! = 120$$ $$5!+3 / 3 = 123 / 3 =41$$correct $$5!+4 / 4 = 124 / 4 =31$$ correct $$5!+5 / 5 = 125 / 5 =25$$ correct $$5!+6 / 6 = 126 / 6 =21$$correct $$5!+7 / 7 = 127 / 7 =18.14$$INCORRECT !!!! so...? what am i doing wrong??? You replaced 15! by 5!. Thus you can not factor out 7 out of 5!+8 and this is the whole point here. did it on excel and seems like you are right..... 15! 15!+x (15!+x)/x 1,307,674,368,000.00 1,307,674,368,003.00 435,891,456,001.00 1,307,674,368,000.00 1,307,674,368,004.00 326,918,592,001.00 1,307,674,368,000.00 1,307,674,368,005.00 261,534,873,601.00 1,307,674,368,000.00 1,307,674,368,006.00 217,945,728,001.00 1,307,674,368,000.00 1,307,674,368,007.00 186,810,624,001.00 1,307,674,368,000.00 1,307,674,368,008.00 163,459,296,001.00 now have to go back and understand how it works.....using the example taht I used (which you mentioned I should use smaller numbers), how do you know when x is too big for a and b, when k!+a<x<k!+b ? Math Expert V Joined: 02 Sep 2009 Posts: 64317 Re: factor factorials [#permalink] ### Show Tags 1 2 zisis wrote: Bunuel wrote: zisis wrote: I am able to follow until the point that I have highlighted red....... tried to work your point my self so decided to find if x for 5!+3<x<5!+8 works with your statement above.... so - $$5! = 120$$ $$5!+3 / 3 = 123 / 3 =41$$correct $$5!+4 / 4 = 124 / 4 =31$$ correct $$5!+5 / 5 = 125 / 5 =25$$ correct $$5!+6 / 6 = 126 / 6 =21$$correct $$5!+7 / 7 = 127 / 7 =18.14$$INCORRECT !!!! so...? what am i doing wrong??? You replaced 15! by 5!. Thus you can not factor out 7 out of 5!+8 and this is the whole point here. did it on excel and seems like you are right..... 15! 15!+x (15!+x)/x 1,307,674,368,000.00 1,307,674,368,003.00 435,891,456,001.00 1,307,674,368,000.00 1,307,674,368,004.00 326,918,592,001.00 1,307,674,368,000.00 1,307,674,368,005.00 261,534,873,601.00 1,307,674,368,000.00 1,307,674,368,006.00 217,945,728,001.00 1,307,674,368,000.00 1,307,674,368,007.00 186,810,624,001.00 1,307,674,368,000.00 1,307,674,368,008.00 163,459,296,001.00 now have to go back and understand how it works.....using the example taht I used (which you mentioned I should use smaller numbers), how do you know when x is too big for a and b, when k!+a<x<k!+b ? It seems that you don't understand the explanation. No need to check in excel: no integer $$x$$ satisfying $$15!+2\leq{x}\leq{15!+15}$$ will be a prime number. There are 14 numbers satisfying it: If $$x=15!+2$$ then we can factor out 2, so $$x$$ would be multiple of 2, thus not a prime; If $$x=15!+3$$ then we can factor out 3, so $$x$$ would be multiple of 3, thus not a prime; ... If $$x=15!+15$$ then we can factor out 15, so $$x$$ would be multiple of 15, thus not a prime. Also 15!+{any multiple of prime less than or equal to 13} also won't be a prime number as we can factor out this prime (15! has all primes less than or equal to 13). Now, for $$x=15!+1$$ or $$x=15!+17$$ or $$x=15!+19$$ we can not say for sure whether they are primes or not. In fact they are such a huge numbers that without a computer it's very hard and time consuming to varify their primality. Hope it's clear. _________________ Manager  Joined: 20 Oct 2013 Posts: 50 Re: If x is an integer, does x have a factor n such that 1 < [#permalink] ### Show Tags good qs!! and explainaton bunnel i am stalking you on GMATclub.com Intern  S Joined: 06 Nov 2016 Posts: 25 Re: If x is an integer, does x have a factor n such that 1 < [#permalink] ### Show Tags Bunuel wrote: zisis wrote: If x is an integer, does x have a factor n such that 1 < n < x? (1) x > 3! (2) 15! + 2 ≤ x ≤ 15! + 15 I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............ If x is an integer, does x have a factor n such that 1 < n < x? Question basically asks: is $$x$$ a prime number? If it is, then it won't have a factor $$n$$ such that $$1<n<x$$ (definition of a prime number). (1) $$x>3!$$ --> $$x$$ is more than some number (3!). $$x$$ may or may not be a prime. Not sufficient. (2) $$15!+2\leq{x}\leq{15!+15}$$ --> $$x$$ can not be a prime. For instance if $$x=15!+8=8*(2*3*4*5*6*7*9*10*11*12*13*14*15+1)$$, then $$x$$ is a multiple of 8, so not a prime. Same for all other numbers in this range: $$x=15!+k$$, where $$2\leq{k}\leq{15}$$ will definitely be a multiple of $$k$$ (as weould be able to factor out $$k$$ out of $$15!+k$$). Sufficient. Answer: B. Instead of multiplying all the numbers from 1 to 15, why not just take prime numbers between 1 and 15. That should still work, right? Math Expert V Joined: 02 Sep 2009 Posts: 64317 Re: If x is an integer, does x have a factor n such that 1 < [#permalink] ### Show Tags SVP482 wrote: Bunuel wrote: zisis wrote: If x is an integer, does x have a factor n such that 1 < n < x? (1) x > 3! (2) 15! + 2 ≤ x ≤ 15! + 15 I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............ If x is an integer, does x have a factor n such that 1 < n < x? Question basically asks: is $$x$$ a prime number? If it is, then it won't have a factor $$n$$ such that $$1<n<x$$ (definition of a prime number). (1) $$x>3!$$ --> $$x$$ is more than some number (3!). $$x$$ may or may not be a prime. Not sufficient. (2) $$15!+2\leq{x}\leq{15!+15}$$ --> $$x$$ can not be a prime. For instance if $$x=15!+8=8*(2*3*4*5*6*7*9*10*11*12*13*14*15+1)$$, then $$x$$ is a multiple of 8, so not a prime. Same for all other numbers in this range: $$x=15!+k$$, where $$2\leq{k}\leq{15}$$ will definitely be a multiple of $$k$$ (as weould be able to factor out $$k$$ out of $$15!+k$$). Sufficient. Answer: B. Instead of multiplying all the numbers from 1 to 15, why not just take prime numbers between 1 and 15. That should still work, right? No, it won't. Why should it? There are primes as well as non-primes from 1 to 15. _________________ Intern  B Joined: 28 Dec 2010 Posts: 21 If x is an integer, does x have a factor n such that 1 < [#permalink] ### Show Tags 1 Given: x is an integer Question: Does x has a factor n such that 1<n<x? --> This a 'yes' or 'no' question If x is not a prime number, then n (factor of x) should always be less equal than x. Example: x = 8, factors of x i.e. n can be 1,2,4,8. If x is a prime number, then n (factor of x) should always be 1, x Example: x = 3, factors of x i.e n can be 1, 3 Statement 1: x > 4! => 4*3*2*1 = 24 now factors of x can be 1, 2, 3, 4, 6, 8, 12, 24. For the values of 1,2,3,4,6,12 it satisfies the condition we get 'yes' for the question but for 24 we get 'no'. Therefore, it is (not sufficient). Statement 2: 15! + 2 ≤ x ≤ 15! + 15 --> range of value for x 15! = 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 15! + 2 = 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 + 2 = 2(15*14*13*12*11*10*9*8*7*6*5*4*3*1 + 1) ---> is completely divisible by 2 (2 is a the factor of 15!+2) 15! +15 =15(14*13*12*11*10*9*8*7*6*5*4*3*2*1 + 1) ---> is completely divisible by 15 (15 is a the factor15!+15) From here we can see that for any value of x in the range it's factor will only be less than x. Therefore, we can definately say "yes" that n will be less than x. (Sufficient) Intern  Joined: 28 Feb 2015 Posts: 1 Re: If x is an integer, does x have a factor n such that 1 < [#permalink] ### Show Tags 1 Bunuel wrote: zisis wrote: If x is an integer, does x have a factor n such that 1 < n < x? (1) x > 3! (2) 15! + 2 ≤ x ≤ 15! + 15 I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............ If x is an integer, does x have a factor n such that 1 < n < x? Question basically asks: is $$x$$ a prime number? If it is, then it won't have a factor $$n$$ such that $$1<n<x$$ (definition of a prime number). (1) $$x>3!$$ --> $$x$$ is more than some number (3!). $$x$$ may or may not be a prime. Not sufficient. (2) $$15!+2\leq{x}\leq{15!+15}$$ --> $$x$$ cannot be a prime. For instance if $$x=15!+8=8*(2*3*4*5*6*7*9*10*11*12*13*14*15+1)$$, then $$x$$ is a multiple of 8, so not a prime. Same for all other numbers in this range: $$x=15!+k$$, where $$2\leq{k}\leq{15}$$ will definitely be a multiple of $$k$$ (as weould be able to factor out $$k$$ out of $$15!+k$$). Sufficient. Answer: B. Hey Bunuel, regarding statement 2: Let's say we would add a prime number bigger than 15 to the 15!, e.g. 17. Because it is not possible to factor out 17, the statement is not sufficient, right? Because the result might be prime or not. On the other hand, if we would add 16, we could still factor out 2 and 8 from 15! correct? Best regards chalmers15 Math Expert V Joined: 02 Sep 2009 Posts: 64317 Re: If x is an integer, does x have a factor n such that 1 < [#permalink] ### Show Tags chalmers15 wrote: Bunuel wrote: zisis wrote: If x is an integer, does x have a factor n such that 1 < n < x? (1) x > 3! (2) 15! + 2 ≤ x ≤ 15! + 15 I am sure i ve seen a quesiton like this one before, but tot forgot how to solve it............ If x is an integer, does x have a factor n such that 1 < n < x? Question basically asks: is $$x$$ a prime number? If it is, then it won't have a factor $$n$$ such that $$1<n<x$$ (definition of a prime number). (1) $$x>3!$$ --> $$x$$ is more than some number (3!). $$x$$ may or may not be a prime. Not sufficient. (2) $$15!+2\leq{x}\leq{15!+15}$$ --> $$x$$ cannot be a prime. For instance if $$x=15!+8=8*(2*3*4*5*6*7*9*10*11*12*13*14*15+1)$$, then $$x$$ is a multiple of 8, so not a prime. Same for all other numbers in this range: $$x=15!+k$$, where $$2\leq{k}\leq{15}$$ will definitely be a multiple of $$k$$ (as weould be able to factor out $$k$$ out of $$15!+k$$). Sufficient. Answer: B. Hey Bunuel, regarding statement 2: Let's say we would add a prime number bigger than 15 to the 15!, e.g. 17. Because it is not possible to factor out 17, the statement is not sufficient, right? Because the result might be prime or not. On the other hand, if we would add 16, we could still factor out 2 and 8 from 15! correct? Best regards chalmers15 You are almost right. Yes, 15! + 16 is not a prime because we can factor out an integer greater than 1 from 15! + 16. Yes, 15! + 17 might be a prime as well as not. But 15! + 17 is some specific number, so theoretically we could find out whether it's prime or not. If it were a prime, then $$15!+2\leq{x}\leq{15!+17}$$ would not be sufficient and if it were NOT a prime, then $$15!+2\leq{x}\leq{15!+17}$$ would be sufficient. Turns out that 15! + 17 is NOT a prime number: 15! + 17 = 1307674368017 = 9157×142805981. So, if (2) were $$15!+2\leq{x}\leq{15!+17}$$ it would still be sufficient. But you won't get such number on the GMAT because you'll need a computer to check whether such a large number is a prime (there are some particular values of $$x=n!+1$$ 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.). _________________ Intern  B Joined: 04 Aug 2018 Posts: 7 Re: If x is an integer, does x have a factor n such that 1 < [#permalink] ### Show Tags VeritasPrepBrian wrote: I love this question, so I'll chime in even if only to bump it to the top so more people see it! Hopefully most can see pretty quickly that x > 3! just means x > 6, and that isn't nearly enough to tell us whether it is prime. Statement 2 is pretty neat, though: 15! + 2 ≤ x ≤ 15! + 15 Think about 15!. 15! is 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1, which means that: 15! is a multiple of 2 15! is a multiple of 3 15! is a multiple of 4 etc.... 15! is a multiple of 15 Now think about multiples of 2. Every SECOND number is a multiple of 2: 2, 4, 6, 8, 10. You create multiples of 2 by adding 2 to a previous multiple of 2. If a number is even, adding 2 "keeps it" even. The same holds for 3. Every THIRD number (3, 6, 9, 12, 15, 18) is a multiple of 3. If you have a multiple of 3 and add 3 to it, it's still a multiple of 3. This will hold for all of these numbers - because 15! is a multiple of every number between 2 and 15, then adding any number in that range of 2-15 will ensure that the new number remains a multiple of that number, and the new number will not be prime. Therefore, statement 2 is sufficient, and the correct answer is B. Now...consider the number 15! + 1. It's too big a number to know offhand whether it's prime, but we do know that it is NOT a multiple of any numbers 2-15. In order to be a multiple of 2, we'd have to add a multiple of 2 to 15!, and 1 breaks us off of that every-second-number cycle. Same for 3 - we'd need to add a multiple of 3 in order to keep 15! on that every-third cycle, so 15! is not a multiple of 3. We can prove that 15! + 1 is not divisible by any numbers between 2 and 15. It's largest prime factor must then be 17 or greater. Understanding that ideology and being able to determine divisibility of large numbers can be quite helpful on questions that might otherwise seem impossible. Thanks for posting this question! I have a question, if 15! + 1 is a prime number ? 15! + 1=1(15*14*13*.....*2+1) so 15! + 1 is a prime number? Veritas Prep Representative G Joined: 26 Jul 2010 Posts: 425 If x is an integer, does x have a factor n such that 1 < [#permalink] ### Show Tags 1 Top Contributor 1 xiaoxueren wrote: I have a question, if 15! + 1 is a prime number ? 15! + 1=1(15*14*13*.....*2+1) so 15! + 1 is a prime number? Yeah really good question. There *is* of course an answer to that, but not one you can prove in 2-3 minutes without a calculator so the GMAT couldn't really ask you about that. You *can* prove that it's not divisible by anything between 2 and 15, but you don't have any easy proof for prime numbers greater than 15 (could it be divisible by 17 or 31 or 53? We just don't know). Consider 5! + 1. 5! is 120 so 5! + 1 = 121. We know that 5! + 1 can't be divisible by 2, 3, 4, or 5, but the factorial setup doesn't give us any insight as to whether it's divisible by 11 (it is). So with 15! + 1, we can't predict (without a calculator at least) which larger prime factors it might have, so in the world of GMAT we can't prove whether it's prime or not. _________________ Brian Curriculum Developer, Instructor, and Host of Veritas Prep On Demand Save$100 on live Veritas Prep GMAT Courses and Admissions Consulting

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Re: If x is an integer, does x have a factor n such that 1 <  [#permalink]

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VeritasPrepBrian wrote:
I love this question, so I'll chime in even if only to bump it to the top so more people see it!

Hopefully most can see pretty quickly that x > 3! just means x > 6, and that isn't nearly enough to tell us whether it is prime.

Statement 2 is pretty neat, though: 15! + 2 ≤ x ≤ 15! + 15

Think about 15!. 15! is 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1, which means that:

15! is a multiple of 2
15! is a multiple of 3
15! is a multiple of 4
etc....
15! is a multiple of 15

Now think about multiples of 2. Every SECOND number is a multiple of 2: 2, 4, 6, 8, 10. You create multiples of 2 by adding 2 to a previous multiple of 2. If a number is even, adding 2 "keeps it" even.

The same holds for 3. Every THIRD number (3, 6, 9, 12, 15, 18) is a multiple of 3. If you have a multiple of 3 and add 3 to it, it's still a multiple of 3.

This will hold for all of these numbers - because 15! is a multiple of every number between 2 and 15, then adding any number in that range of 2-15 will ensure that the new number remains a multiple of that number, and the new number will not be prime.

Therefore, statement 2 is sufficient, and the correct answer is B.

Now...consider the number 15! + 1. It's too big a number to know offhand whether it's prime, but we do know that it is NOT a multiple of any numbers 2-15. In order to be a multiple of 2, we'd have to add a multiple of 2 to 15!, and 1 breaks us off of that every-second-number cycle. Same for 3 - we'd need to add a multiple of 3 in order to keep 15! on that every-third cycle, so 15! is not a multiple of 3. We can prove that 15! + 1 is not divisible by any numbers between 2 and 15. It's largest prime factor must then be 17 or greater.

Understanding that ideology and being able to determine divisibility of large numbers can be quite helpful on questions that might otherwise seem impossible. Thanks for posting this question!

Just loved it..Thanks.

So this is how I applied it it for a small number.

4! + X here X can lie between 2 and 4 So, 4! + X will have factor from 4! to be keep it on the cycle of X

In short 4! + 2 cannot be prime since 4!+2 = 4*3*2*1+ 2=26 (26 factors as 2 )
Similarly 4! + 3 or 4!+ as long as the added term (except 1) is inside the factorial
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Re: If x is an integer, does x have a factor n such that 1 <  [#permalink]

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VeritasPrepBrian wrote:
xiaoxueren wrote:

I have a question, if 15! + 1 is a prime number ?
15! + 1=1(15*14*13*.....*2+1)
so 15! + 1 is a prime number?

Yeah really good question. There *is* of course an answer to that, but not one you can prove in 2-3 minutes without a calculator so the GMAT couldn't really ask you about that. You *can* prove that it's not divisible by anything between 2 and 15, but you don't have any easy proof for prime numbers greater than 15 (could it be divisible by 17 or 31 or 53? We just don't know).

Consider 5! + 1. 5! is 120 so 5! + 1 = 121. We know that 5! + 1 can't be divisible by 2, 3, 4, or 5, but the factorial setup doesn't give us any insight as to whether it's divisible by 11 (it is).

So with 15! + 1, we can't predict (without a calculator at least) which larger prime factors it might have, so in the world of GMAT we can't prove whether it's prime or not.

Really appreciate, Now I understand this question. Non-Human User Joined: 09 Sep 2013
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Re: If x is an integer, does x have a factor n such that 1 <  [#permalink]

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_________________ Re: If x is an integer, does x have a factor n such that 1 <   [#permalink] 22 Oct 2019, 05:06

# If x is an integer, does x have a factor n such that 1 <  