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M28-52

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New post 16 Sep 2014, 01:44
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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 M28-52  [#permalink]

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New post 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: M28-52  [#permalink]

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New post 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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M28-52  [#permalink]

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New post 28 Feb 2017, 12:14
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: M28-52  [#permalink]

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New post 28 Feb 2017, 21:22
1
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: M28-52  [#permalink]

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New post 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: M28-52  [#permalink]

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New post 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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Re: M28-52  [#permalink]

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New post 22 Jun 2017, 03:39
alexlovesgmat wrote:
With all respect
As it is mentioned factorial of a negative is undefined. Then how can the -17 ! be in scope?


Where do you see factorial of negative number there?

For (2) we only know that c cannot be negative but you cannot use info from (1) when considering (2).
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Re: M28-52  [#permalink]

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New post 22 Jun 2017, 04:36
Omg thanks bunuel
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Re: M28-52  [#permalink]

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New post 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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Re: M28-52  [#permalink]

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New post 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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Re: M28-52  [#permalink]

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New post 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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Re: M28-52  [#permalink]

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New post 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: M28-52  [#permalink]

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New post 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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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.


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Re: M28-52 &nbs [#permalink] 21 Jul 2018, 12:18
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