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

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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 $\boxed{B}$ .

@@ Line 1: / Line 1: @@
+==Problem 12==
+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>
+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>?
+<math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15 </math>
+==Solution==
+===First iteration===
+To get <math>f(k) = 27</math>, you could either have <math>f(27 - 3)</math> and add <math>3</math>, or <math>f(27\cdot 2)</math> and divide by <math>2</math>.
+If you had the former, you would have <math>f(24)</math>, and the function's rule would have you divide.  Thus, <math>k=54</math> is the only number for which <math>f(k) = 27</math>.
+===Second iteration===
+Going out one step, if you have <math>f(f(k)) = 27</math>, you would have to have <math>f(k) = 54</math>.   For <math>f(k) = 54</math>, you would either have <math>f(54-3)</math> and add <math>3</math>, or <math>f(54\cdot 2)</math> and divide by <math>2</math>.
+Both are possible:  <math>f(51)</math> and <math>f(108)</math> return values of <math>54</math>.  Thus, <math>f(f(51)) = f(54) = 27</math>, and <math>f(f(108)) = f(54) = 27</math>.
+===Third iteration===
+Going out the final step, if you have <math>f(f(f(k))) = 27</math>, you would have to have <math>f(f(k))) = 51</math> or <math>f(f(k)) = 108</math>.
+If you doubled either of these, <math>k</math> would not be odd.  So you must subtract <math>3</math>.
+If you subtract <math>3</math> from <math>51</math>, you would compute <math>f(48)</math>, which would halve it, and not add the <math>3</math> back.
+If you subtract <math>3</math> from <math>108</math>, you would compute <math>f(105)</math>, which would add the <math>3</math> back.
+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>.
 ==See also==
 {{AHSME box|year=1996|num-b=11|num-a=13}}

1996 AHSME (Problems • Answer Key • Resources)
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