Factorials
Find the sum of the possible integer values of x such that
( ( ( x ! ) ! ) ! ) … = x .
The answer is 3.
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2 solutions
Solution:
x can be expressed as:
x ! ! ! ! ! ! ! . . . ) / ( ( x − 1 ) ! ! ! ! ! ! ! ! ! . . ) = ( x ! ! ! ! ! ! ! ! ! . . . )
1 = ( x − 1 ) ! ! ! ! ! ! ! ! ! . . )
But 1 = 1 ! ! ! ! ! ! ! . . . or 0 ! ! ! ! ! ! ! ! ! . . .
So, we can say that: x = 1 + 1 or 0 + 1 which is 1 a n d 2 ..
Therefore, S U M of all possible x is: 1 + 2 = 3 A n s .
The only way to preserve x!!!!!!!!... = x is when x! = x. Since x! = x, x!/x = 1. That can be written as x(x-1)...(1)/x, which means (x-1)! = 1. This is true when x=2 or x=1. So the answer is 2+1 = 3. However, (infinity)!!!!!!!!... = infinity is also true, so technically the answer could also be 2+1+infinity = infinity. I gave my answer as 3 and got this right, but I think the question should probably stipulate that x <> infinity.