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# M28-52

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Math Expert
Joined: 02 Sep 2009
Posts: 46297

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16 Sep 2014, 01:44
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Difficulty:

55% (hard)

Question Stats:

63% (00:59) correct 38% (01:45) wrong based on 40 sessions

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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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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.

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Joined: 14 Jul 2014
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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.

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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Joined: 17 Aug 2016
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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.

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: 46297

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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.

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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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.

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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Joined: 04 Sep 2016
Posts: 19

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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?
Math Expert
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Posts: 46297

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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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22 Jun 2017, 04:36
Omg thanks bunuel
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Joined: 13 Mar 2017
Posts: 610
Location: India
Concentration: General Management, Entrepreneurship
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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.

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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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CAT 99th percentiler : VA 97.27 | DI-LR 96.84 | QA 98.04 | OA 98.95
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.

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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
Re: M28-52   [#permalink] 11 Mar 2018, 13:33
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# M28-52

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