Let \(f(n) = \begin{cases} n^2+1 & \text{if }n\text{ is odd} \\ \dfrac{n}{2} & \text{if }n\text{ is even} \end{cases}.\)

For how many integers n from 1 to 100, inclusive, does \(f ( f (\dotsb f (n) \dotsb )) = 1 \) for some number of applications of f?

HelpBot Sep 22, 2020

#3**+1 **

I played with this on a spreadsheet for quite a while.

But this is that I ended up with

f(2)=1 n=2 works and for integers from 1 to 100 this is the only n that will give an output of 1

If f(n)=2 then this will work too. So n can be 4 f(4)=2 f(2)=1 so f(f(4))=1

by the same logic f(8) f(16) f(32) and f(64) will all work.

so n can be 1,2,4,8,16,32 or 64

So I get 7.

I have broken my own guideline and given you a final answer, I hope you learn from it.

Melody Sep 23, 2020