A function f from the integers to the integers is defined as follows: f(n)= n+3 & if n is odd, (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?

1996 AMC 12 Problem 12

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12.

A function

f

from the integers to the integers is defined as follows:

f(n)= \begin{cases} n+3 & \text{if \(n\) is odd},\\ \frac{n}{2} & \text{if \(n\) is even}. \end{cases}

Suppose

k

is odd and

f(f(f(k)))=27.

What is the sum of the digits of

k?

3

6

9

12

15

Because

k

is odd, the first iterate is

k+3

, which is even

Work backward from

27

, checking the parity required by each inverse branch

Since

k

is odd,

f(k)=k+3

is even, so

f(f(k))=\frac{k+3}{2}.

If this value were odd, adding

3

could not produce

27

because that would require the value

24.

Therefore it is even and is halved to

27,

\frac{k+3}{2}=54.

Hence

k=105,

whose digits sum to

6.

The correct answer is B.