Difference between revisions of "1996 AHSME Problems/Problem 12"
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| + | ==Problem 12== | ||
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| + | A function <math>f</math> from the integers to the integers is defined as follows: | ||
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| + | <cmath> f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ \dfrac{n}{2} &\text{if n is even}\end{cases} </cmath> | ||
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| + | 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>? | ||
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| + | <math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15 </math> | ||
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| + | ==Solution== | ||
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| + | ===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>. | ||
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| + | 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>. | ||
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| + | 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>. | ||
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| + | |||
| + | ===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>. | ||
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| + | If you doubled either of these, <math>k</math> would not be odd. So you must subtract <math>3</math>. | ||
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| + | 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. | ||
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| + | 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. | ||
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| + | 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>. | ||
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| + | ==Solution 2 (rigorous but easy)== | ||
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| + | <cmath> f(f(f(k))) = 27 </cmath> | ||
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| + | We will work from the inside to the outside and alternate \( k \) between even and odd. | ||
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| + | If \( k \) is odd, then \( f(k) \) in terms of \( k \) is \(\frac{k+3}{2}\). | ||
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| + | Next, we have \( f\left(\frac{k+3}{2}\right) \), and \(\frac{k+3}{2}\) is even, so \( f\left(\frac{k+3}{2}\right) = \frac{k+3}{4} \). | ||
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| + | We are given that \( f\left(\frac{k+3}{4}\right) = 27 \). | ||
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| + | <cmath>\frac{k+3}{4} = 27 \implies k = 105.</cmath> | ||
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| + | The sum of its digits is \( 1 + 0 + 5 = 6 \). | ||
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| + | {gnv12} | ||
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==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=11|num-a=13}} | {{AHSME box|year=1996|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:54, 23 August 2024
Contents
Problem 12
A function 
from the integers to the integers is defined as follows:
![\[f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ \dfrac{n}{2} &\text{if n is even}\end{cases}\]](http://latex.artofproblemsolving.com/7/3/d/73dfb1825c6300ca81dea05ed375b9a49b6698ca.png)
Suppose 
is odd and 
. What is the sum of the digits of ?
Solution
First iteration
To get 
, you could either have 
and add 
, or 
and divide by 
.
If you had the former, you would have 
, and the function's rule would have you divide. Thus, 
is the only number for which .
Second iteration
Going out one step, if you have 
, you would have to have 
. For , you would either have 
and add , or 
and divide by .
Both are possible: 
and 
return values of 
. Thus, 
, and 
.
Third iteration
Going out the final step, if you have , you would have to have 
or 
.
If you doubled either of these, would not be odd. So you must subtract .
If you subtract from 
, you would compute 
, which would halve it, and not add the back.
If you subtract from 
, you would compute 
, which would add the back.
Thus, 
, and 
is odd. The desired sum of the digits is 
, and the answer is 
.
Solution 2 (rigorous but easy)
![\[f(f(f(k))) = 27\]](http://latex.artofproblemsolving.com/1/d/a/1da68e7f255b946a11a288004d2e63ee7d1d2fc5.png)
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 \( f\left(\frac{k+3}{4}\right) = 27 \).
![\[\frac{k+3}{4} = 27 \implies k = 105.\]](http://latex.artofproblemsolving.com/b/2/1/b2145409bf834ca657aa73ec0ef2abe807c892df.png)
The sum of its digits is \( 1 + 0 + 5 = 6 \).
{gnv12}
See also
| 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 | ||
| All AHSME Problems and Solutions | ||
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