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Re M3001 [#permalink]
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16 Sep 2014, 01:44
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Official Solution: If \(x\) is a positive integer greater than 1, is \(x\) a prime number? (1) \(x\) does not have a factor p such that \(2 < p < x\). Notice that all odd primes satisfy this statement as well as integer 4 (4 does not have a factor p such that \(2 < p < 4\)). Not sufficient. (2) The product of any two factors of \(x\) is greater than 2 but less than 10. This implies that \(x\) can be 3, 5, or 7. Sufficient. Notice that \(x\) cannot be an even number because any even number has 1 and 2 as its factors and the product of these factors is 2, not greater than 2 as given in the statement. Also notice that x cannot be 9 because 3 and 9 both are factors of 9 and 3*9=27>10. Answer: B
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Re: M3001 [#permalink]
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25 Nov 2016, 07:34
Bunuel wrote: Official Solution:
If \(x\) is a positive integer greater than 1, is \(x\) a prime number? (1) \(x\) does not have a factor p such that \(2 < p < x\). Notice that all odd primes satisfy this statement as well as integer 4 (4 does not have a factor p such that \(2 < p < 4\)). Not sufficient. (2) The product of any two factors of \(x\) is greater than 2 but less than 10. This implies that \(x\) can be 3, 5, or 7. Sufficient. Notice that \(x\) cannot be an even number because any even number has 1 and 2 as its factors and the product of these factors is 2, not greater than 2 as given in the statement. Also notice that x cannot be 9 because 3 and 9 both are factors of 9 and 3*9=27>10.
Answer: B Could you explain the S2 in detail? I could not understand. If product of any 2 factors is less than 10; then they can be 1*9 which means x is not a prime.



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Re: M3001 [#permalink]
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25 Nov 2016, 08:03
chismooo wrote: Bunuel wrote: Official Solution:
If \(x\) is a positive integer greater than 1, is \(x\) a prime number? (1) \(x\) does not have a factor p such that \(2 < p < x\). Notice that all odd primes satisfy this statement as well as integer 4 (4 does not have a factor p such that \(2 < p < 4\)). Not sufficient. (2) The product of any two factors of \(x\) is greater than 2 but less than 10. This implies that \(x\) can be 3, 5, or 7. Sufficient. Notice that \(x\) cannot be an even number because any even number has 1 and 2 as its factors and the product of these factors is 2, not greater than 2 as given in the statement. Also notice that x cannot be 9 because 3 and 9 both are factors of 9 and 3*9=27>10.
Answer: B Could you explain the S2 in detail? I could not understand. If product of any 2 factors is less than 10; then they can be 1*9 which means x is not a prime. Hi Statement II talks of any two factors, but you are just looking at two factors here ... So if you are looking at 9.. Factors are 1,3,9.. So if you pick 3 and 9, the product is 3*9=27.. 1*3 and 1*9 may be less than 10 but we are looking at any two Hope it helps
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Re: M3001 [#permalink]
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04 Mar 2017, 21:57
Hi, RE Statement 2, couldn't X be 4 (which would give it factors of 1, 2 and 4?), giving it options 3, 4, 5 and 7, making it insufficient since it gives both prime and nonprime numbers?



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Re: M3001 [#permalink]
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05 Mar 2017, 03:55
JLim90 wrote: Hi, RE Statement 2, couldn't X be 4 (which would give it factors of 1, 2 and 4?), giving it options 3, 4, 5 and 7, making it insufficient since it gives both prime and nonprime numbers? No it could not be 4 or any other even number. Please read the solution again: If \(x\) is a positive integer greater than 1, is \(x\) a prime number? (1) \(x\) does not have a factor p such that \(2 < p < x\). Notice that all odd primes satisfy this statement as well as integer 4 (4 does not have a factor p such that \(2 < p < 4\)). Not sufficient. (2) The product of any two factors of \(x\) is greater than 2 but less than 10. This implies that \(x\) can be 3, 5, or 7. Sufficient. Notice that \(x\) cannot be an even number because any even number has 1 and 2 as its factors and the product of these factors is 2, not greater than 2 as given in the statement. Also notice that x cannot be 9 because 3 and 9 both are factors of 9 and 3*9=27>10. Answer: B
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Re: M3001 [#permalink]
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04 Jun 2017, 09:49
if x = 6, then product of its factors will be 2 x 3 = 6 which satisfies the statement but x is not prime. kindly clarify



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Re: M3001 [#permalink]
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04 Jun 2017, 10:53
kupratik1 wrote: if x = 6, then product of its factors will be 2 x 3 = 6 which satisfies the statement but x is not prime. kindly clarify Hi The Statement 2 states that product of any two factors of x is greater than 2 but less than 10. This means that no matter which two factors of x you take, their product should not exceed or be equal to 10. Also, no matter which two factors of x you take, their product should not be less than or equal to 2. So x=6 will NOT satisfy this property. because I can state two such factors of 6 (3 and 6) which multiply to give product greater than 10. Similarly, x=4 will also NOT satisfy this property, because I can state two factors of 4 (1 and 2) which multiply to give a product equal to 2. The required number x should be such that we should not be able to come up with any two factors of x, which multiply to give 2 or less.... NOR should we be able to come up with any two factors of x which multiply to give 10 or more.



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Re: M3001 [#permalink]
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06 Jun 2017, 05:59
Hello Sir,
I have great difficulty in understanding S1. Could you please provide a detailed explanation?
Thank you.



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Re: M3001 [#permalink]
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06 Jun 2017, 09:33
Manoraaju wrote: Hello Sir,
I have great difficulty in understanding S1. Could you please provide a detailed explanation?
Thank you. (1) says that if you primefactorize x, it won't have a prime which is greater than 2 but less than x itself. Prime numbers, generally don't have factors which are greater than 1 and less than this prime itself (from definition of primes: a prime number is a positive integer with exactly two factors 1 and itself). So, any odd prime will satisfy this condition. But not only odd primes, 4 also does not have a prime which is greater than 2 but less than 4 itself (factors of 4 are 1, 2, and 4). So, odd primes as well as 4 satisfy (1), which makes this statement not sufficient. Hope it's clear.
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Re: M3001 [#permalink]
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06 Jun 2017, 19:11
Bunuel wrote: Manoraaju wrote: Hello Sir,
I have great difficulty in understanding S1. Could you please provide a detailed explanation?
Thank you. (1) says that if you primefactorize x, it won't have a prime which is greater than 2 but less than x itself. Prime numbers, generally don't have factors which are greater than 1 and less than this prime itself (from definition of primes: a prime number is a positive integer with exactly two factors 1 and itself). So, any odd prime will satisfy this condition. But not only odd primes, 4 also does not have a prime which is greater than 2 but less than 4 itself (factors of 4 are 1, 2, and 4). So, odd primes as well as 4 satisfy (1), which makes this statement not sufficient. Hope it's clear. Omg, crystal clear. Thank you very much Mr. Bunuel.



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Bunuel wrote: Official Solution:
If \(x\) is a positive integer greater than 1, is \(x\) a prime number? (1) \(x\) does not have a factor p such that \(2 < p < x\). Notice that all odd primes satisfy this statement as well as integer 4 (4 does not have a factor p such that \(2 < p < 4\)). Not sufficient. (2) The product of any two factors of \(x\) is greater than 2 but less than 10. This implies that \(x\) can be 3, 5, or 7. Sufficient. Notice that \(x\) cannot be an even number because any even number has 1 and 2 as its factors and the product of these factors is 2, not greater than 2 as given in the statement. Also notice that x cannot be 9 because 3 and 9 both are factors of 9 and 3*9=27>10.
Answer: B Dear Bunuel, In Fact 1, can I say the following: P =5, x= 7........Answer to the question is Yes P =5, x= 14......Answer to question is No why did you restrict the example to 4? Also, it is not clear what you mean by factor? is it prime factorization or factors of a number? Can you please elaborate? Thanks



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Re M3001 [#permalink]
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10 Aug 2017, 22:34
I think this is a poorquality question. What about 4 for option 2. 4 has 1,2, 4 as factors. 2*4 = 8 <10&>2



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Re: M3001 [#permalink]
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10 Aug 2017, 23:33



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Re: M3001 [#permalink]
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15 Sep 2017, 18:59
This is a good question and the explanation is clear. Thank you!



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Re: M3001 [#permalink]
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14 Nov 2017, 20:10
I think this is a high quality question and it has an excellent answer! kudos!










