Question

A function ' f ' from integers to integers is defined as follows: f(x) = n+3 if n is odd and n/2 if n is even. Suppose k is odd and f(f(f(k))) = 27. What is the sum of the digits of k?

• A. 3
• B. 6
• C. 9
• D. 8

Answer 1

The correct option is B 6

Option (b)

Since k is odd then f(f(f(k))) =27
$\Rightarrow$ f(f(k+3)) = 27------------(1) since k is odd , k+3 will be even.
Hence, expanding (1) we get, $f\left(\frac{k+3}{2}\right)=27$--------------(2)
Here $\frac{k+3}{2}$ can either be even or odd

Case 1:
Let $\frac{k+3}{2}$ is even
Then (2) reduces to $\frac{k+3}{4}=27\Rightarrow k=105$ here sum of digits is 6

Case 2:
Let $\frac{k+3}{2}$ is odd
Then (2) reduces to $\frac{k+3}{2}+3=27\Rightarrow k=45$

Substituting in $\frac{k+3}{2}$ it comes to be 24 which is not odd hence there is a contradiction hence
k = 105 here sum of digits is 6 is the only solution.

Hence, choice (b) is the correct answer.