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Devil's Dozen!!!
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27 Oct 2018, 14:08
Hi Bunuel! I am a bit confused for the case n=1. If n=1 then statement one equates to 5 and statement 2 equates to 6. Hence the possible values of P are 5,2,3. none of which is a factor of 1? If each statement is true, then P should then be a prime factor of 5 and either 2 or 3, but that's impossible? Will be grateful if you could clarify. Thank you. Bunuel wrote: 2. If n is a positive integer and p is a prime number, is p a factor of n!?
(1) p is a factor of (n+2)!n! > if \(n=2\) then \((n+2)!n!=22\) and for \(p=2\) then answer will be YES but for \(p=11\) the answer will be NO. Not sufficient.
(2) p is a factor of (n+2)!/n! > \(\frac{(n+2)!}{n!}=(n+1)(n+2)\) > if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) then answer will be YES but for \(p=3\) the answer will be NO. Not sufficient.
(1)+(2) \((n+2)!n!=n!((n+1)(n+2)1)\). Now, \((n+1)(n+2)1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are coprime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)1\), thus in order \(p\) to be a factor of \(n!*((n+1)(n+2)1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.
Answer: C.



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Re: Devil's Dozen!!!
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28 Oct 2018, 01:31
gmat800live wrote: Hi Bunuel! I am a bit confused for the case n=1. If n=1 then statement one equates to 5 and statement 2 equates to 6. Hence the possible values of P are 5,2,3. none of which is a factor of 1? If each statement is true, then P should then be a prime factor of 5 and either 2 or 3, but that's impossible? Will be grateful if you could clarify. Thank you. Bunuel wrote: 2. If n is a positive integer and p is a prime number, is p a factor of n!?
(1) p is a factor of (n+2)!n! > if \(n=2\) then \((n+2)!n!=22\) and for \(p=2\) then answer will be YES but for \(p=11\) the answer will be NO. Not sufficient.
(2) p is a factor of (n+2)!/n! > \(\frac{(n+2)!}{n!}=(n+1)(n+2)\) > if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) then answer will be YES but for \(p=3\) the answer will be NO. Not sufficient.
(1)+(2) \((n+2)!n!=n!((n+1)(n+2)1)\). Now, \((n+1)(n+2)1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are coprime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)1\), thus in order \(p\) to be a factor of \(n!*((n+1)(n+2)1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.
Answer: C. When we consider two statements together then n cannot be 1 because p cannot be a prime which is simultaneous a factor of 5 and 6.
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Re: Devil's Dozen!!!
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21 Feb 2019, 03:52
Hi, If someone could explain the 13th question in more detail it would be helpful
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Give +1 kudos if this answer helps..!!



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Re: Devil's Dozen!!!
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31 Jul 2019, 02:06
Selection of n numbers from K numbers is kCn, isn't it? And of kCn ways, only one way will be such that all numbers are in ascending order. Therefore, the probability is = 1/kCn Shouldn't answer be C? Bunuel wrote: 7. Set A consists of k distinct numbers. If n numbers are selected from the set onebyone, where n<=k, what is the probability that numbers will be selected in ascending order?
(1) Set A consists of 12 even consecutive integers; (2) n=5.
We should understand following two things: 1. The probability of selecting any n numbers from the set is the same. Why should any subset of n numbers have higher or lower probability of being selected than some other subset of n numbers? Probability doesn't favor any particular subset.
2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1<x_2<...<x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these n! ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\).
Hence, according to the above the only thing we need to know to answer the question is the size of the subset (n) we are selecting from set A.
Answer: B.



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Re: Devil's Dozen!!!
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31 Jul 2019, 02:12
gvij2017 wrote: Selection of n numbers from K numbers is kCn, isn't it? And of kCn ways, only one way will be such that all numbers are in ascending order. Therefore, the probability is = 1/kCn Shouldn't answer be C? Bunuel wrote: 7. Set A consists of k distinct numbers. If n numbers are selected from the set onebyone, where n<=k, what is the probability that numbers will be selected in ascending order?
(1) Set A consists of 12 even consecutive integers; (2) n=5.
We should understand following two things: 1. The probability of selecting any n numbers from the set is the same. Why should any subset of n numbers have higher or lower probability of being selected than some other subset of n numbers? Probability doesn't favor any particular subset.
2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1<x_2<...<x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these n! ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\).
Hence, according to the above the only thing we need to know to answer the question is the size of the subset (n) we are selecting from set A.
Answer: B. I tried to elaborate a bit here: https://gmatclub.com/forum/m27184482.html
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Re: Devil's Dozen!!!
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12 Oct 2019, 16:03
Bunuel wrote: 10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?
Quite tricky.
(1) There are total 99 snakes in Pandora's box. Clearly insufficient.
(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.
Answer: B. What if there are two pairs of VC VC, then you would have two cobras? Question is asking for absolute number of cobras, but you could have infinite pair of vipre cobras right? thus E.. Maybe I am missing something.. thank you



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Re: Devil's Dozen!!!
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13 Oct 2019, 00:48
DanielMx wrote: Bunuel wrote: 10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?
Quite tricky.
(1) There are total 99 snakes in Pandora's box. Clearly insufficient.
(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.
Answer: B. What if there are two pairs of VC VC, then you would have two cobras? Question is asking for absolute number of cobras, but you could have infinite pair of vipre cobras right? thus E.. Maybe I am missing something.. thank you We cannot have more than 1 cobra. If there are 2 cobras and 2 vipers then the second statement will NOT hold. (2) says from ANY two snakes from Pandora's box at least one is a viper. If there are 2 cobras and 2 vipers then we could have two snakes from which BOTH are cobras.
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Re: Devil's Dozen!!!
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