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If x is an integer, does x have a factor n such that 1 < [#permalink]
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08 Sep 2010, 11:52
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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............
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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.
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Re: factor factorials [#permalink]
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08 Sep 2010, 13:17
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



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Re: factor factorials [#permalink]
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08 Sep 2010, 17:54
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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 215 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 215. 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 everysecondnumber cycle. Same for 3  we'd need to add a multiple of 3 in order to keep 15! on that everythird 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!
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Re: factor factorials [#permalink]
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09 Sep 2010, 06:26
Nice rephrase. Somehow in option 1 I took my values as 3!, 4!, 5! and thought they all will have a factor as needed.
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Re: factor factorials [#permalink]
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09 Sep 2010, 14:28
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???



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Re: factor factorials [#permalink]
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Re: factor factorials [#permalink]
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09 Sep 2010, 15:49
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 ?



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Re: factor factorials [#permalink]
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09 Sep 2010, 16:19
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.
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Re: If x is an integer, does x have a factor n such that 1 < [#permalink]
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21 Feb 2017, 06:33
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?



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Re: If x is an integer, does x have a factor n such that 1 < [#permalink]
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21 Feb 2017, 10:55
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 nonprimes from 1 to 15.
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