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D01-20

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Re D01-20  [#permalink]

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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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Re: D01-20  [#permalink]

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New post 09 Sep 2017, 15:07
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Re: D01-20  [#permalink]

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New post 13 Mar 2018, 17:48
I got it wrong in the tests due to oversight on my part.

Option 1 - The factorial of a number can be odd only if its 0 or 1. This gives a value of b. But C can be any number greater than b. - Not sufficient.

Option 2 - c! is a prime number. The only number whose factorial is prime is 2. So we get the value of c as 2.
but we do not have value of a and b. Not sufficient.

Considering both the options, we can find that the numbers are consecutive numbers.

Ans: C
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D01-20  [#permalink]

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New post 02 Aug 2019, 22:14
all integers; a<b<c

consec?

1. Median of a! b! c! is odd
a=0= 0! = 1
b=1 = 1! = 1
c=2 = 2! = 2
consec

OR

a=0 =0! = 1
b=1 = 1! = 1
c= 6 = 6!= 720

not consec

Therefore statement 1 is insufficient

Statement 2
c! = prime
c= 2!

But a can still equal 1
a= -1
b= 1
c=2 =2! thus not consecutive

since the only constraint for a and b is a<b<c then we can make a and b equal anything.

Combining the two statements, combined, a!<b!<c!

A factorial is defined only for non-negative integer numbers.

Thus 0!<1!<2! is the only possible solution within the constraint.

Sufficient.
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Re D01-20  [#permalink]

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New post 21 Sep 2019, 00:06
I think this is a high-quality question and I agree with explanation. Great question! Tests a mix of multiple basic facts on numbers.
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Re: D01-20  [#permalink]

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New post 16 Oct 2019, 00:17
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


I don't quite agree with the solution here. Using option (2) , I come up with the number c =2 , since none of the numbers can be negative, we are left with only 0 and 1, here in the problem statement , it is given that a<b<c , hence a has to take 0 and b has to take 1
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Re: D01-20  [#permalink]

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New post 16 Oct 2019, 00:22
dasoisheretorule 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


I don't quite agree with the solution here. Using option (2) , I come up with the number c =2 , since none of the numbers can be negative, we are left with only 0 and 1, here in the problem statement , it is given that a<b<c , hence a has to take 0 and b has to take 1


I think that's addressed on the previous pages. Please re-read. Hope it helps.
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Re: D01-20  [#permalink]

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New post 16 Oct 2019, 00:29
dasoisheretorule 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


I don't quite agree with the solution here. Using option (2) , I come up with the number c =2 , since none of the numbers can be negative, we are left with only 0 and 1, here in the problem statement , it is given that a<b<c , hence a has to take 0 and b has to take 1




I see my mistake now. Until I use (1) , I can't come up with the information that a and b are also supposed to be within the set of positive integers. Thanks [color=#ff0000]Bunnel [/color]
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Re: D01-20  [#permalink]

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New post 17 Nov 2019, 06:17
Just a short question: since we know that factorials for negative numbers is undefined should't it be safe to assume that a,b,c are positive ?
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Re: D01-20  [#permalink]

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New post 17 Nov 2019, 07:40
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Re: D01-20  [#permalink]

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New post 22 Nov 2019, 16:07
bunuel
why not you consider values of negative integer for a?
for ex. a=-1,-2,-3.. these numbers also satisfy given condition
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Re: D01-20  [#permalink]

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New post 22 Nov 2019, 19:16
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Re: D01-20   [#permalink] 22 Nov 2019, 19:16

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