January 17, 2019 January 17, 2019 08:00 AM PST 09:00 AM PST Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL. January 19, 2019 January 19, 2019 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52230

Question Stats:
54% (01:17) correct 46% (01:29) wrong based on 157 sessions
HideShow timer Statistics
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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Math Expert
Joined: 02 Sep 2009
Posts: 52230

Re M2852
[#permalink]
Show Tags
16 Sep 2014, 00:44
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
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 14 Jul 2014
Posts: 93

Re: M2852
[#permalink]
Show Tags
12 Jan 2015, 04:10
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" ?



Intern
Joined: 17 Aug 2016
Posts: 48

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?



Math Expert
Joined: 02 Sep 2009
Posts: 52230

Re: M2852
[#permalink]
Show Tags
28 Feb 2017, 20:22
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?
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 17 Aug 2016
Posts: 48

Re: M2852
[#permalink]
Show Tags
04 Mar 2017, 12:10
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.



Intern
Joined: 04 Sep 2016
Posts: 19

Re: M2852
[#permalink]
Show Tags
22 Jun 2017, 02:24
With all respect As it is mentioned factorial of a negative is undefined. Then how can the 17 ! be in scope?



Math Expert
Joined: 02 Sep 2009
Posts: 52230

Re: M2852
[#permalink]
Show Tags
22 Jun 2017, 02:39



Intern
Joined: 04 Sep 2016
Posts: 19

Re: M2852
[#permalink]
Show Tags
22 Jun 2017, 03:36
Omg thanks bunuel



Director
Joined: 13 Mar 2017
Posts: 696
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: M2852
[#permalink]
Show Tags
22 Jun 2017, 22:43
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.
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
MBA Social Network : WebMaggu
Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".



Director
Joined: 13 Mar 2017
Posts: 696
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: M2852
[#permalink]
Show Tags
22 Jun 2017, 22:46
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.
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
MBA Social Network : WebMaggu
Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".



Manager
Joined: 26 Feb 2018
Posts: 79
Location: United Arab Emirates
GMAT 1: 710 Q47 V41 GMAT 2: 770 Q49 V47

Re: M2852
[#permalink]
Show Tags
11 Mar 2018, 12:33
Does anyone know if the GMAT tests that 0! is 1? I've never seen this on an OG question



Intern
Joined: 31 Jan 2018
Posts: 1

Re: M2852
[#permalink]
Show Tags
21 Jul 2018, 09:43
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?



Math Expert
Joined: 02 Sep 2009
Posts: 52230

Re: M2852
[#permalink]
Show Tags
21 Jul 2018, 11:18
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).
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 07 Oct 2014
Posts: 5

Re: M2852
[#permalink]
Show Tags
18 Dec 2018, 22:01
Can someone please rewrite the question as it's hard to tell what the numbers represent in this format... thanks



Math Expert
Joined: 02 Sep 2009
Posts: 52230

Re: M2852
[#permalink]
Show Tags
18 Dec 2018, 22:17










