Difference between revisions of "2005 AMC 8 Problems/Problem 24"
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<cmath>200, 100, 50, 25, 24, 12, 6, 3, 2, 1</cmath> | <cmath>200, 100, 50, 25, 24, 12, 6, 3, 2, 1</cmath> | ||
| − | This took | + | This took keystrokes. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=23|num-a=25}} | {{AMC8 box|year=2005|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:34, 26 October 2016
Problem
A certain calculator has only two keys [+1] and [x2]. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed "9" and you pressed [+1], it would display "10." If you then pressed [x2], it would display "20." Starting with the display "1," what is the fewest number of keystrokes you would need to reach "200"?
Solution
It is easier to work backwards from 
, and have keys that display [-1] and [x0.5]. Use the second key when the number is even, and the first key when the number is odd until you get one. We get:
![\[200, 100, 50, 25, 24, 12, 6, 3, 2, 1\]](http://latex.artofproblemsolving.com/6/4/5/64564a7f73a49ceedaa4fb1c4e023ad5731b1339.png)
This took keystrokes.
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
