New Concept

Algebra Level 3

Let defined the concept of n ! n^{!} .

Definition of n ! n^{!} .

Let n W n \in \mathbb{W} , we defined n ! n^{!} using the formula

n ! = n n 1 n 2 n 3 . . . . 3 2 1 n^{!} = n^{n-1^{n-2^{n-3^{....^{3^{2^{1}}}}}}}

Find n n . If ( n 1 ) ! = 1 (n-1)^{!} = 1

3 2 1 4

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2 solutions

Paul Ryan Longhas
Mar 26, 2015

( n 1 ) ! = 1 = > n 1 = 1 (n-1)^{!} = 1 => n-1 = 1 or ( n 2 ) ! = 0 (n-2)^{!} = 0

n 1 = 1 = > n = 2 n-1 = 1 => n = 2 or ( n 2 ) ! = 0 = > n 2 = 0 = > n = 2 (n-2)^! = 0 => n-2 = 0 => n = 2

= > n = 2 => n=2

Alex Delhumeau
Jun 8, 2015

In this case, the solution is simply the trivial case where n 1 = 1 n = 2 n-1=1\therefore n=2 . Nice problem, though!

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