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08 Sep 2010, 11:52
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
70% (00:59) correct
30%
(01:43)
wrong
based on 1560
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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
(1) x > 3!
(2) 15! + 2 ≤ x ≤ 15! + 15
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 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!
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!
08 Sep 2010, 12:01
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zisis
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.
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