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avatar+399 

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?

 Sep 22, 2020
 #3
avatar+111120 
+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.

 Sep 23, 2020
 #4
avatar+399 
+1

Thank you that does make sense. I will solve more problems in this category

 Sep 23, 2020

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