+0  
 
0
83
8
avatar+809 

A function \(f\) from the integers to the integers is defined as follows:
Suppose \(k\) is odd and \(f(f(f(k))) = 27.\) Find \(k\).

 

Do we have to work backwards?

 Dec 27, 2018
 #1
avatar+98107 
+2

Mmmm....I haven't seen one like this.....but.....here's my best shot....

 

Let us suppose that the function is

 

f(k)  = 3k        and let k = 1

 

So

 

f(1) =  3 (1)  = 3

f ( f(1) )  = f(3)  =   3(3) =  9

f ( f ( f ( 1 ) )  ) =   f(9)  =  3(9) =   27

 

So..... k =  1

 

Anyone else have other thoughts???

 

cool cool cool

 Dec 27, 2018
edited by CPhill  Dec 27, 2018
 #2
avatar+34 
+1

Hmm, this looks very tricky.

Since \(k\) is odd, we denote \(f(k)=k+3.\) Any odd number plus \(3\) is even, so \(f(k + 3) = \frac{k + 3}{2}\).

But, if \(\frac{k + 3}{2}\) is odd, therefore \(f \left( \frac{k + 3}{2} \right) = \frac{k + 3}{2} + 3 = 27.\) Then this leads for \(k=45\), and \(f(f(f(45))) = f(f(48)) = f(24) = 12\), so it is even. \(f \left( \frac{k + 3}{2} \right) = \frac{k + 3}{4} = 27\) leads to \(105\), so this is your answer?

 

Check: \(f(f(f(105))) = f(f(108)) = f(54) = 27.\)

.
 Dec 27, 2018
 #3
avatar+17331 
+1

Wondering if    f(x) = x          then f(27) = 27

f(f(27) ) = 27

f(f(f(27))) = 27     then k = 27 (which is odd)......    Hmmmmm

 Dec 27, 2018
 #4
avatar+98107 
+1

Everyone has their own take on this.....I'd like to see the "real" solution, myself.....LOL!!!

 

 

cool cool cool

CPhill  Dec 27, 2018
 #5
avatar+34 
+1

Yes, we have to see.

neworleans06  Dec 27, 2018
 #8
avatar+142 
0

LOL. You really got me with that pun there. Lmao

noobieatmath  Dec 28, 2018
 #6
avatar+809 
+1

Looks like the correct answer is 105, I'll try to come up with a solution on why.

 Dec 27, 2018
 #7
avatar+98107 
0

OK....looks like neworleans wins....!!!!

 

 

cool cool cool

CPhill  Dec 27, 2018

14 Online Users

avatar
avatar
avatar
avatar