Difference between revisions of "1996 AHSME Problems/Problem 12"

Latest revision as of 18:54, 23 August 2024

Problem 12

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

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

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 $\boxed{B}$ .

Solution 2 (rigorous but easy)

$f(f(f(k))) = 27$

We will work from the inside to the outside and alternate $ k $ between even and odd.

If $ k $ is odd, then $ f(k) $ in terms of $ k $ is $\frac{k+3}{2}$.

Next, we have $ f\left(\frac{k+3}{2}\right) $, and $\frac{k+3}{2}$ is even, so $ f\left(\frac{k+3}{2}\right) = \frac{k+3}{4} $.

We are given that $ f\left(\frac{k+3}{4}\right) = 27 $.

$\frac{k+3}{4} = 27 \implies k = 105.$

The sum of its digits is $ 1 + 0 + 5 = 6 $.

{gnv12}

@@ Line 3: / Line 3: @@
 A function <math>f</math> from the integers to the integers is defined as follows:
-<cmath> f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ n/2 &\text{if n is even}\end{cases} </cmath>
+<cmath> f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ \dfrac{n}{2} &\text{if n is even}\end{cases} </cmath>
 Suppose <math>k</math> is odd and <math>f(f(f(k))) = 27</math>.  What is the sum of the digits of <math>k</math>?
@@ Line 32: / Line 32: @@
 Thus, <math>f(f(f(105))) = f(f(108)) = f(54) = 27</math>, and <math>105</math> is odd.  The desired sum of the digits is <math>6</math>, and the answer is <math>\boxed{B}</math>.
+==Solution 2 (rigorous but easy)==
+<cmath> f(f(f(k))) = 27 </cmath>
+We will work from the inside to the outside and alternate \( k \) between even and odd.
+If \( k \) is odd, then \( f(k) \) in terms of \( k \) is \(\frac{k+3}{2}\).
+Next, we have \( f\left(\frac{k+3}{2}\right) \), and \(\frac{k+3}{2}\) is even, so \( f\left(\frac{k+3}{2}\right) = \frac{k+3}{4} \).
+We are given that \( f\left(\frac{k+3}{4}\right) = 27 \).
+<cmath>\frac{k+3}{4} = 27 \implies k = 105.</cmath>
+The sum of its digits is \( 1 + 0 + 5 = 6 \).
+{gnv12}
 ==See also==
 {{AHSME box|year=1996|num-b=11|num-a=13}}
 {{MAA Notice}}

1996 AHSME (Problems • Answer Key • Resources)
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