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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............

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.

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! _________________

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....

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....

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....

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 ?

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....

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.

Re: If x is an integer, does x have a factor n such that 1 < [#permalink]
02 Oct 2013, 10:02

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