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If y is a positive integer, is y prime?
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Updated on: 24 Oct 2012, 05:45
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If y is a positive integer, is y prime? (1) y>4! (2) 11!12<y<11!2
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Originally posted by LM on 24 Oct 2012, 05:02.
Last edited by Bunuel on 24 Oct 2012, 05:45, edited 1 time in total.
Edited the question.



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Re: If y is a positive integer, is y prime?
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24 Oct 2012, 05:22
LM wrote: If y is a positive integer, is y prime? 1.\(y>4!\) 2.\(11!12<y<11!2\) (1) obviously not sufficient. There are many primes greater than 4! as well as nonprimes. (2) y can be one of the integers 11!  11, 11!  10, 11!  9, ... , 11!  2, 11!  3. It is easy to see that all the numbers on the above list are certainly not primes. 11! = 2x3x4x5x6x7x8x9x10x11, and between 11! and the term subtracted, there is in each case a common factor. So, the first number on the list is divisible by 11, the second by 10,..., the last number is divisible by 3. Sufficient. Answer B.
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Re: If y is a positive integer, is y prime?
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24 Oct 2012, 05:47
LM wrote: If y is a positive integer, is y prime?
(1) y>4! (2) 11!12<y<11!2 Similar question to practice: 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. Answer: B. Discussed here: doestheintegerkhaveafactorpsuchthat1pk126735.htmlHope it helps.
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Re: If y is a positive integer, is y prime?
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24 Oct 2012, 05:48
LM wrote: If y is a positive integer, is y prime?
(1) y>4! (2) 11!12<y<11!2 Another similar question: 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. Discussed here: ifxisanintegerdoesxhaveafactornsuchthat100670.htmlHope it helps.
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Re: If y is a positive integer, is y prime?
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26 Oct 2012, 09:03
EvaJager wrote: LM wrote: If y is a positive integer, is y prime? 1.\(y>4!\) 2.\(11!12<y<11!2\) (1) obviously not sufficient. There are many primes greater than 4! as well as nonprimes. (2) y can be one of the integers 11!  11, 11!  10, 11!  9, ... , 11!  2, 11!  3. It is easy to see that all the numbers on the above list are certainly not primes. 11! = 2x3x4x5x6x7x8x9x10x11, and between 11! and the term subtracted, there is in each case a common factor. So, the first number on the list is divisible by 11, the second by 10,..., the last number is divisible by 3. Sufficient. Answer B. Hi Eva, can you help to explain me why B is the answer? you mentioned that, "y can be one of the integers 11!  11, 11!  10, 11!  9, ... , 11!  2, 11!  3." and "the first number on the list is divisible by 11, the second by 10,..., the last number is divisible by 3", can you help to discuss it in more detail? thanks.
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Re: If y is a positive integer, is y prime?
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26 Oct 2012, 09:19
ss58146 wrote: EvaJager wrote: LM wrote: If y is a positive integer, is y prime? 1.\(y>4!\) 2.\(11!12<y<11!2\) (1) obviously not sufficient. There are many primes greater than 4! as well as nonprimes. (2) y can be one of the integers 11!  11, 11!  10, 11!  9, ... , 11!  2, 11!  3. It is easy to see that all the numbers on the above list are certainly not primes. 11! = 2x3x4x5x6x7x8x9x10x11, and between 11! and the term subtracted, there is in each case a common factor. So, the first number on the list is divisible by 11, the second by 10,..., the last number is divisible by 3. Sufficient. Answer B. Hi Eva, can you help to explain me why B is the answer? you mentioned that, "y can be one of the integers 11!  11, 11!  10, 11!  9, ... , 11!  2, 11!  3." and "the first number on the list is divisible by 11, the second by 10,..., the last number is divisible by 3", can you help to discuss it in more detail? thanks. The integers between x  12 and x  2 are x  11, x  10, ..., x  3. In our case x =11!. Take common factor between 11! and the term that is subtracted from it: For example, 11!  11 = 2x3x4x5x6x7x8x9x10x11  11 = 11(2x3x4x5x6x7x8x9x10  1) is divisible by 11, so it is not a prime, as the number in the parentheses is greater than 1.
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Re: If y is a positive integer, is y prime?
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26 Oct 2012, 09:42
it is not a prime, as the number in the parentheses is greater than 1. I get it now .. Thank you..
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If y is a positive integer, is y prime?
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04 Apr 2014, 23:04
If y is a positive integer, is y prime?
1) y > 4! 2) 11! – 12 < y < 11! – 2



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Re: If y is a positive integer, is y prime?
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05 Apr 2014, 02:40



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Re: If y is a positive integer, is y prime?
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20 Apr 2018, 13:47
LM wrote: If y is a positive integer, is y prime?
(1) y > 4! (2) 11!  12 < y < 11!  2 Target question: is y prime? Statement 1: y > 4! In other words, y > 24 This does not help us determine whether or not y is prime. Consider these two conflicting cases: Case a: y = 29, in which case y IS primeCase b: y = 25, in which case y is NOT primeSince we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Statement 2: 11! – 12 < y < 11! – 2 Let's examine a few possible values for y. y = 11! – 11y = ( 11)(10)(9)....(5)(4)(3)(2)(1)  11 y = 11[(10)(9)....(5)(4)(3)(1)  1] Since y is a multiple of 11, y is NOT primey = 11! – 10y = (11)( 10)(9)....(5)(4)(3)(2)(1)  10 y = 10[(11)(9)....(5)(4)(3)(1)  1] Since y is a multiple of 10, y is NOT primey = 11! – 9y = (11)(10)( 9)....(5)(4)(3)(2)(1)  9 y = 9[(11)(10)....(5)(4)(3)(1)  1] Since y is a multiple of 9, y is NOT primeAs you can see, this pattern can be repeated all the way up to y = 11!  1. In EVERY case, y is NOT primeSince we can answer the target question with certainty, statement 2 is SUFFICIENT Answer: B Cheers, Brent
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