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D01-20

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Re: D01-20 [#permalink]

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New post 10 Dec 2016, 11:21
This is a little confusing. If factorial is not defined for negative numbers, then a and b must be positive.

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Re: D01-20 [#permalink]

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New post 17 Feb 2017, 06:34
Bunuel
As per the solution given:
(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.

it says b is 0 or 1. Is 0 an odd number?

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New post 17 Feb 2017, 09:05
Darkhorse12 wrote:
Bunuel
As per the solution given:
(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.

it says b is 0 or 1. Is 0 an odd number?


0 is an even integer. But 0! = 1 = odd.
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Re: D01-20 [#permalink]

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New post 22 Apr 2017, 21:29
How is 0 factorial 1?

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New post 10 May 2017, 11:41
How can we determine that c must be 2 (not any other prime number) from the :

2) c! is a prime number.

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New post 10 May 2017, 11:53
Albert__ wrote:
How can we determine that c must be 2 (not any other prime number) from the :

2) c! is a prime number.


n! = 1*2*3*...*n

Now, if c is any other number than 2, c! will have at least 2 and 3 as its factor and we know that a prime number has only two factors 1 and itself, thus for c! to be prime c must be 2. For example, if c = 4, then c! = 4! = 1*2*3*4, which is not a prime.

Hope it's clear.
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New post 10 May 2017, 11:56
My bad... stupid mistake, thanks anyway !

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New post 15 Jun 2017, 07:05
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.


Love your explanation.

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Re: D01-20 [#permalink]

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New post 07 Aug 2017, 05:52
In this question, what we know from statement A is that b! is odd. So b could either be 0 or 1.

If b is 0, then a can still assume the value 0. a does not have to be negative.

The second statement tells us that c is 2, but gives no details on a and b.

Combining both the statements still gives us three possible values for {a!,b!,c!}:

I: {0!,0!,2!}
II: {0!,1!,2!}
III: {1!,1!,2!}

I and III would mean that the answer is NO and II would mean the answer is YES.

Going by this logic, I went with E.

Can anyone please correct me and explain how the answer is C? I do not understand any of the given explanations.

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New post 07 Aug 2017, 16:50
anirudhb94 wrote:
In this question, what we know from statement A is that b! is odd. So b could either be 0 or 1.

If b is 0, then a can still assume the value 0. a does not have to be negative.

The second statement tells us that c is 2, but gives no details on a and b.

Combining both the statements still gives us three possible values for {a!,b!,c!}:

I: {0!,0!,2!}
II: {0!,1!,2!}
III: {1!,1!,2!}

I and III would mean that the answer is NO and II would mean the answer is YES.

Going by this logic, I went with E.

Can anyone please correct me and explain how the answer is C? I do not understand any of the given explanations.


Notice that we are told that a < b < c, so I and III are not possible, which leaves only a=0, b=1 and c=2.
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New post 09 Aug 2017, 00:47
Factorial of a number is prime
Can it be anything elses besides 2!?
Or 2! is the only factorial of a number is prime?

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Factorial of a number is prime
Can it be anything elses besides 2!?
Or 2! is the only factorial of a number is prime?


No, it can only be 2. A prime number has only two factors 1 and itself, so factorial of any number but 2 cannot be a prime.
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New post 09 Sep 2017, 15:00
I don't agree with the explanation. i think A should be the answer because a set (1,1,c!) will have a median of 1 always na

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New post 09 Sep 2017, 15:07
samism wrote:
I don't agree with the explanation. i think A should be the answer because a set (1,1,c!) will have a median of 1 always na


You should read solutions and the following discussions more carefully.

The question asks whether a, b, and c are consecutive integers. For (1): a=0, b=1 and c=2 gives an YES answer while a=0, b=1 and c=3 gives a NO answer.
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Re D01-20 [#permalink]

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New post 13 Oct 2017, 12:18
I think this is a high-quality question and I agree with explanation.

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New post 19 Nov 2017, 08:13
I think this is a high-quality question and I agree with explanation.

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New post 10 Dec 2017, 14:18
I think this is a poor-quality question. for (1), you cannot assume the set is provided in increasing order...

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New post 10 Dec 2017, 21:00

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Re: D01-20   [#permalink] 10 Dec 2017, 21:00

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