Is nonnegative integer x odd? (1) x! = |x - 1|! (2) x! = (x + 1)!

Question

Is nonnegative integer x odd?

(1) x! = |x - 1|!
(2) x! = (x + 1)!

RMD007 · Answer

X is a non negative integer, so x = 0 or x>0
We know that 0! = 1 and 1! = 1

Statement 1:

If x= 0 0! = |-1|! = 1 
If x = 1 1!= 0!= 1

X can be 0 or 1. Not sufficient.

Statement 2:
If x = 0, 0! = 1! = 1
If x = 1, 1! is not equal to 2! so only possible value for x is 0. Sufficient.

B.

yezz · Answer

X is integer >= 0

from 1

x! = /x-1/! , least value of x! = 1 and x! = (x-1)! when x>=1 and X! = (1-x)! when x<1 (i.e. x= 0)

first case for x>=1 , x* (x-1)! = (x-1)! thus (x-1)!*x = 0 and since x>=1 thus x =1
2nd case when x = 0 .... insuff

from 2

x! = (x+1)! only possible way for this to hold is x = 0 .... suff

B

AkshdeepS · Answer

We know  x\geq{0}

We also know 0! = 1 & 1! = 1

As per St 1:
So If x = 0, 0! = |0-1|! or 1 = 1 (x = 0 = even)
 If x = 1, 1! = |1-1|! or 1 = 0! or 1 = 1 So (x = 1 = odd) (Not sufficient)

As per St 2:
 This statement is valid only when x = 0; So (x = 0 = even) (Sufficient)

(B)

parthwagh1234 · Answer

This question presents a great starting point for discussion, but it would greatly benefit from additional perspectives and insights.

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