Question

A function $f$ from integers to integers is defined as $f\left(x\right)=\{\begin{array}{cc}n+3,& nϵodd\\ \frac{n}{2}& nϵeven\end{array}$ . Suppose $kϵ$ odd and $f\left(f\right(f\left(k\right)\left)\right)=27$, then the sum of digits of $k$ is

• A. $3$
• B. $6$
• C. $9$
• D. $12$

Answer 1

The correct option is B $6$

Since k is odd,
$f\left(k\right)=k+3$
Now $k+3$ is even.
Hence
$f\left(f\right(k\left)\right)=\frac{k+3}{2}$.
Now $k+3$ is always divisible by 4 for all $kϵoddnumber$.
Hence $\frac{k+3}{2}$ is even.
Then
$f\left(f\right(f\left(k\right)\left)\right)=\frac{k+3}{4}$
$=27$
$k+3=108$
$k=105$
Hence sum of the digits is 6.