A function f: Z→Z is defined as fn= n+1 n∈odd integer n/2 n∈even integer . If k∈odd integer and fffk=33, then the sum of the digits of k is

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Correct answer: (B) 5. ∵ k∈odd integer fk=k+1 which is even ∴ffk=(k+1)/2 If (k+1)/2 is odd ⇒33=(k+1)/2+1⇒k=63 not possible (because (k+1)/2 is even) If (k+1)/2 is even ⇒33=(k+1)/22⇒33=(k+1)/4⇒k=131 Sum of digits =5

A function $f : Z \to Z$ is defined as $f (n) = \{\begin{array}{l} n + 1 & n \in odd integer \\ \frac{n}{2} & n \in even integer \end{array} .$ If $k \in odd integer$ and $f (f (f (k))) = 33,$ then the sum of the digits of $k$ is

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A function f: Z→Z is defined as fn=n+1n∈odd integern2n∈even integer. If k∈odd integer and fffk=33, then the sum of the digits of k is

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A function $f : Z \to Z$ is defined as $f (n) = \{\begin{array}{l} n + 1 & n \in odd integer \\ \frac{n}{2} & n \in even integer \end{array} .$ If $k \in odd integer$ and $f (f (f (k))) = 33,$ then the sum of the digits of $k$ is