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# Piecewise?

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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
+100439
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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???

Dec 27, 2018
edited by CPhill  Dec 27, 2018
#2
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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
+18293
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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
+100439
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Everyone has their own take on this.....I'd like to see the "real" solution, myself.....LOL!!!

CPhill  Dec 27, 2018
#5
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Yes, we have to see.

neworleans06  Dec 27, 2018
#8
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LOL. You really got me with that pun there. Lmao

noobieatmath  Dec 28, 2018
#6
+814
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Looks like the correct answer is 105, I'll try to come up with a solution on why.

Dec 27, 2018
#7
+100439
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OK....looks like neworleans wins....!!!!

CPhill  Dec 27, 2018