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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.
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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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.
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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