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If \(a\), \(b\), and \(c\) are integers and \(a \lt b \lt c\), are \(a\), \(b\), and \(c\) consecutive integers? (1) The median of \(\{a!, \ b!, \ c!\}\) is an odd number. (2) \(c!\) is a prime number.
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Re M2852
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16 Sep 2014, 01:44
Official Solution: Note that: A. The factorial of a negative number is undefined. B. \(0!=1\). C. Only two factorials are odd: \(0!=1\) and \(1!=1\). D. Factorial of a number which is prime is \(2!=2\). (1) The median of \(\{a!, \ b!, \ c!\}\) is an odd number. This implies that \(b!=odd\). Thus \(b\) is 0 or 1. But if \(b=0\), then \(a\) is a negative number, so in this case \(a!\) is not defined. Therefore \(a=0\) and \(b=1\), so the set is \(\{0!, \ 1!, \ c!\}=\{1, \ 1, \ c!\}\). Now, if \(c=2\), then the answer is YES but if \(c\) is any other number then the answer is NO. Not sufficient. (2) \(c!\) is a prime number. This implies that \(c=2\). Not sufficient. (1)+(2) From above we have that \(a=0\), \(b=1\) and \(c=2\), thus the answer to the question is YES. Sufficient. Answer: C
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Re: M2852
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12 Jan 2015, 05:10
Bunuel wrote: Official Solution:
Note that: A. The factorial of a negative number is undefined. B. \(0!=1\). C. Only two factorials are odd: \(0!=1\) and \(1!=1\). D. Factorial of a number which is prime is \(2!=2\). (1) The median of \(\{a!, \ b!, \ c!\}\) is an odd number. This implies that \(b!=odd\). Thus \(b\) is 0 or 1. But if \(b=0\), then \(a\) is a negative number, so in this case \(a!\) is not defined.Therefore \(a=0\) and \(b=1\), so the set is \(\{0!, \ 1!, \ c!\}=\{1, \ 1, \ c!\}\). Now, if \(c=2\), then the answer is YES but if \(c\) is any other number then the answer is NO. Not sufficient. (2) \(c!\) is a prime number. This implies that \(c=2\). Not sufficient. (1)+(2) From above we have that \(a=0\), \(b=1\) and \(c=2\), thus the answer to the question is YES. Sufficient.
Answer: C Query w.r.t Stmnt 1 I understand the highlighted part  where a! is not defined. But can we officially consider such a scenario where mathematically it is not possible to derive it? Is this the reason that in Stmnt 1 , we only consider the Red part? And the actual answer (Yes/No) depends on the vaue of "C" ?



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Bunuel wrote: Official Solution:
Note that: A. The factorial of a negative number is undefined. B. \(0!=1\). C. Only two factorials are odd: \(0!=1\) and \(1!=1\). D. Factorial of a number which is prime is \(2!=2\). (1) The median of \(\{a!, \ b!, \ c!\}\) is an odd number. This implies that \(b!=odd\). Thus \(b\) is 0 or 1. But if \(b=0\), then \(a\) is a negative number, so in this case \(a!\) is not defined. Therefore \(a=0\) and \(b=1\), so the set is \(\{0!, \ 1!, \ c!\}=\{1, \ 1, \ c!\}\). Now, if \(c=2\), then the answer is YES but if \(c\) is any other number then the answer is NO. Not sufficient. (2) \(c!\) is a prime number. This implies that \(c=2\). Not sufficient. (1)+(2) From above we have that \(a=0\), \(b=1\) and \(c=2\), thus the answer to the question is YES. Sufficient.
Answer: C Hi Bunuel, why c=2 is not enough? if a and b are integers and a<b<c with c = aren't we sure b=1 and a=0? What else could it be? Could you please clarify this?



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Re: M2852
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28 Feb 2017, 21:22
bazu wrote: Bunuel wrote: Official Solution:
Note that: A. The factorial of a negative number is undefined. B. \(0!=1\). C. Only two factorials are odd: \(0!=1\) and \(1!=1\). D. Factorial of a number which is prime is \(2!=2\). (1) The median of \(\{a!, \ b!, \ c!\}\) is an odd number. This implies that \(b!=odd\). Thus \(b\) is 0 or 1. But if \(b=0\), then \(a\) is a negative number, so in this case \(a!\) is not defined. Therefore \(a=0\) and \(b=1\), so the set is \(\{0!, \ 1!, \ c!\}=\{1, \ 1, \ c!\}\). Now, if \(c=2\), then the answer is YES but if \(c\) is any other number then the answer is NO. Not sufficient. (2) \(c!\) is a prime number. This implies that \(c=2\). Not sufficient. (1)+(2) From above we have that \(a=0\), \(b=1\) and \(c=2\), thus the answer to the question is YES. Sufficient.
Answer: C Hi Bunuel, why c=2 is not enough? if a and b are integers and a<b<c with c = aren't we sure b=1 and a=0? What else could it be? Could you please clarify this? Why is it necessary if c=2, a and b to be 0 and 1? Why cannot a be 100 and b be 17?
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Re: M2852
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04 Mar 2017, 13:10
Bunuel wrote: bazu wrote: Bunuel wrote: Official Solution:
Note that: A. The factorial of a negative number is undefined. B. \(0!=1\). C. Only two factorials are odd: \(0!=1\) and \(1!=1\). D. Factorial of a number which is prime is \(2!=2\). (1) The median of \(\{a!, \ b!, \ c!\}\) is an odd number. This implies that \(b!=odd\). Thus \(b\) is 0 or 1. But if \(b=0\), then \(a\) is a negative number, so in this case \(a!\) is not defined. Therefore \(a=0\) and \(b=1\), so the set is \(\{0!, \ 1!, \ c!\}=\{1, \ 1, \ c!\}\). Now, if \(c=2\), then the answer is YES but if \(c\) is any other number then the answer is NO. Not sufficient. (2) \(c!\) is a prime number. This implies that \(c=2\). Not sufficient. (1)+(2) From above we have that \(a=0\), \(b=1\) and \(c=2\), thus the answer to the question is YES. Sufficient.
Answer: C Hi Bunuel, why c=2 is not enough? if a and b are integers and a<b<c with c = aren't we sure b=1 and a=0? What else could it be? Could you please clarify this? Why is it necessary if c=2, a and b to be 0 and 1? Why cannot a be 100 and b be 17? Thanks Bunuel, I hope I was only tired, I don't know why but I had in my mind that a b c where positive integers.



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Re: M2852
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22 Jun 2017, 03:24
With all respect As it is mentioned factorial of a negative is undefined. Then how can the 17 ! be in scope?



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22 Jun 2017, 04:36
Omg thanks bunuel



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Re: M2852
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22 Jun 2017, 23:43
Bunuel wrote: If \(a\), \(b\), and \(c\) are integers and \(a \lt b \lt c\), are \(a\), \(b\), and \(c\) consecutive integers?
(1) The median of \(\{a!, \ b!, \ c!\}\) is an odd number.
(2) \(c!\) is a prime number. Given : a,b,c are integers and a<b<c. DS : What needs to be checked is whether a,b,c are consecutive integers. Now lets start checking the options : (1) median of {a!,b!,c!} is odd number. So b =1. Because if b=0, a will become ve, if b=2, a will become even a<b<c. Hence, a=0, b=1, we don't know anything about c. can be 2,3,4,.... So NOT SUFFICIENT. (2) c! is a prime number. So c can be 2 only because if c=3,4,.... c! wont be a prime number. Now a<b<c and a,b,c are integers So, we know that c=2 , but we can't say anything about a,b can be 1,0,1,2.............. NOT Sufficient Combined : a=0, b=1, c=2. Answer C.
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22 Jun 2017, 23:46
alexlovesgmat wrote: With all respect As it is mentioned factorial of a negative is undefined. Then how can the 17 ! be in scope? By option (2) you can't say anything about a or b. Its just we know that a,d<c i.e.2 So a,b can be 1,0,1,2,3.....................,100,...................., infinity.
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11 Mar 2018, 13:33
Does anyone know if the GMAT tests that 0! is 1? I've never seen this on an OG question



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21 Jul 2018, 10:43
Bunuel wrote: Official Solution:
Note that: A. The factorial of a negative number is undefined. B. \(0!=1\). C. Only two factorials are odd: \(0!=1\) and \(1!=1\). D. Factorial of a number which is prime is \(2!=2\). (1) The median of \(\{a!, \ b!, \ c!\}\) is an odd number. This implies that \(b!=odd\). Thus \(b\) is 0 or 1. But if \(b=0\), then \(a\) is a negative number, so in this case \(a!\) is not defined. Therefore \(a=0\) and \(b=1\), so the set is \(\{0!, \ 1!, \ c!\}=\{1, \ 1, \ c!\}\). Now, if \(c=2\), then the answer is YES but if \(c\) is any other number then the answer is NO. Not sufficient. (2) \(c!\) is a prime number. This implies that \(c=2\). Not sufficient. (1)+(2) From above we have that \(a=0\), \(b=1\) and \(c=2\), thus the answer to the question is YES. Sufficient.
Answer: C Hi Bunuel, From the following Note, Point A: it was derived that in a<b<c, 'a' cannot be negative . For the Condition 1. But if b=0, then a is a negative number, so in this case a! is not defined. If we were to not take Point A in consideration then 'a' can very well be 17! Note that: A. The factorial of a negative number is undefined.B. 0!=10!=1. C. Only two factorials are odd: 0!=10!=1 and 1!=11!=1. D. Factorial of a number which is prime is 2!=22!=2. Using Point A in the Note, why can't the same('a' and 'b' cannot be negative) be derived for Condition 2? Condition 2 states that 'c' is prime. 2! is prime. Therefore, c=2. Since a<b<c, therefore a and b can take 0 and 1 only iff Point A is considered here. So why can't Point A be considered for Condition 2?



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Re: M2852
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21 Jul 2018, 12:18
srishti02 wrote: Bunuel wrote: Official Solution:
Note that: A. The factorial of a negative number is undefined. B. \(0!=1\). C. Only two factorials are odd: \(0!=1\) and \(1!=1\). D. Factorial of a number which is prime is \(2!=2\). (1) The median of \(\{a!, \ b!, \ c!\}\) is an odd number. This implies that \(b!=odd\). Thus \(b\) is 0 or 1. But if \(b=0\), then \(a\) is a negative number, so in this case \(a!\) is not defined. Therefore \(a=0\) and \(b=1\), so the set is \(\{0!, \ 1!, \ c!\}=\{1, \ 1, \ c!\}\). Now, if \(c=2\), then the answer is YES but if \(c\) is any other number then the answer is NO. Not sufficient. (2) \(c!\) is a prime number. This implies that \(c=2\). Not sufficient. (1)+(2) From above we have that \(a=0\), \(b=1\) and \(c=2\), thus the answer to the question is YES. Sufficient.
Answer: C Hi Bunuel, From the following Note, Point A: it was derived that in a<b<c, 'a' cannot be negative . For the Condition 1. But if b=0, then a is a negative number, so in this case a! is not defined. If we were to not take Point A in consideration then 'a' can very well be 17! Note that: A. The factorial of a negative number is undefined.B. 0!=10!=1. C. Only two factorials are odd: 0!=10!=1 and 1!=11!=1. D. Factorial of a number which is prime is 2!=22!=2. Using Point A in the Note, why can't the same('a' and 'b' cannot be negative) be derived for Condition 2? Condition 2 states that 'c' is prime. 2! is prime. Therefore, c=2. Since a<b<c, therefore a and b can take 0 and 1 only iff Point A is considered here. So why can't Point A be considered for Condition 2? For (2) we only know that c cannot be negative but you cannot use info from (1) when considering (2).
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