M28-52 : GMAT Club Tests

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bazu

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Re: M28-52 [#permalink]

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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Bunuel

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Re: M28-52 [#permalink]

28 Feb 2017, 21:22

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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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bazu

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Re: M28-52 [#permalink]

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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alexlovesgmat

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Re: M28-52 [#permalink]

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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Bunuel

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Re: M28-52 [#permalink]

22 Jun 2017, 03:39

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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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alexlovesgmat

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Re: M28-52 [#permalink]

22 Jun 2017, 04:36

Omg thanks bunuel

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Re: M28-52 [#permalink]

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]

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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tinyinthedesert

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Re: M28-52 [#permalink]

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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srishti02

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Re: M28-52 [#permalink]

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]

21 Jul 2018, 12:18

Expert Reply

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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pinguuu

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Re: M28-52 [#permalink]

18 Dec 2018, 23:01

Can someone please rewrite the question as it's hard to tell what the numbers represent in this format... thanks

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Bunuel

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Re: M28-52 [#permalink]

18 Dec 2018, 23:17

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pinguuu wrote:

Can someone please rewrite the question as it's hard to tell what the numbers represent in this format... thanks

Everything looks fine for me. Could you please try refreshing the page? If it does not help, please post a screenshot. Thank you.

PhilipLimantara

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Re: M28-52 [#permalink]

12 Feb 2021, 00:53

Bunuel wrote:

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).

Hi Bunuel,
For statement 2, c! is a prime number and we know from the information initially that a, b, and c are integers and a<b<c. If we are using the common concept that the factorial of a negative number is undefined, why cant we conclude that when C=2, b=1 and a=0? Why do we need to consider the possibility of a negative integer for a and b when the common concept that we know is that the factorial of a negative number is negative?

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Re: M28-52 [#permalink]

30 Aug 2023, 12:29

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Duplicate of D01-20. Unpublished.

gmatclubot

Re: M28-52 [#permalink]

30 Aug 2023, 12:29