Difference between revisions of "1996 AHSME Problems/Problem 12"
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==Solution 2 (rigorous but easy)== | ==Solution 2 (rigorous but easy)== | ||
| − | + | <cmath> f(f(f(k))) = 27 </cmath> | |
| − | We will alternate k between even and odd. | + | We will work from the inside to the outside and alternate \( k \) between even and odd. |
| − | If k is odd, then \ | + | If \( k \) is odd, then \( f(k) \) in terms of \( k \) is \(\frac{k+3}{2}\). |
| − | Next, we have \ | + | 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 \ | + | 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== | ==See also== | ||
{{AHSME box|year=1996|num-b=11|num-a=13}} | {{AHSME box|year=1996|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:16, 30 June 2024
Contents
Problem 12
A function 
from the integers to the integers is defined as follows:
![\[f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ n/2 &\text{if n is even}\end{cases}\]](http://latex.artofproblemsolving.com/2/b/f/2bf71d8b3d5fa6e931eef419bab85379ea2c40ba.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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
