1996 AHSME Problems/Problem 12

Problem 12

A function $f$ from the integers to the integers is defined as follows:

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

Suppose $k$ is odd and $f(f(f(k))) = 27$ . What is the sum of the digits of $k$ ?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$

Solution

First iteration

To get $f(k) = 27$ , you could either have $f(27 - 3)$ and add $3$ , or $f(27\cdot 2)$ and divide by $2$ .

If you had the former, you would have $f(24)$ , and the function's rule would have you divide. Thus, $k=54$ is the only number for which $f(k) = 27$ .

Second iteration

Going out one step, if you have $f(f(k)) = 27$ , you would have to have $f(k) = 54$ . For $f(k) = 54$ , you would either have $f(54-3)$ and add $3$ , or $f(54\cdot 2)$ and divide by $2$ .

Both are possible: $f(51)$ and $f(108)$ return values of $54$ . Thus, $f(f(51)) = f(54) = 27$ , and $f(f(108)) = f(54) = 27$ .

Third iteration

Going out the final step, if you have $f(f(f(k))) = 27$ , you would have to have $f(f(k))) = 51$ or $f(f(k)) = 108$ .

If you doubled either of these, $k$ would not be odd. So you must subtract $3$ .

If you subtract $3$ from $51$ , you would compute $f(48)$ , which would halve it, and not add the $3$ back.

If you subtract $3$ from $108$ , you would compute $f(105)$ , which would add the $3$ back.

Thus, $f(f(f(105))) = f(f(108)) = f(54) = 27$ , and $105$ is odd. The desired sum of the digits is $6$ , and the answer is $\boxed{B}$ .

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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