Difference between revisions of "1996 AHSME Problems/Problem 12"
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Latest revision as of 14:07, 5 July 2013
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 
.
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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