Bunuel wrote:
5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?
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 S2 we get c=2 and we know a<b<c and that a,b,c are non negetive int. So, it's quite evident theat a,b have to be 0,1 ? Hence, sufficient? Am i missing something here ?