Difference between revisions of "2005 AMC 8 Problems/Problem 24"
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| − | First we can start at 200 and work our way down we want to do <math>/2</math> the most so if we come across an odd number we just subtract 1. So <math>200/2</math>=<math>100</math>, <math>100/2</math>=<math>50</math>, <math>50/2</math>=<math>25</math>, <math>25-1</math>=<math>24</math>, <math>24/2</math>=<math>12</math>, <math>12/2</math>=<math>6</math>, <math>6/2</math>=<math>3</math>, <math>3-1</math>=<math>2</math>, and <math>2/2</math>=<math>1</math>. We made our way back to 1 but since it is the amount of times the button is pressed than the answer should be \boxed{\textbf{(B)}\ 9} | + | First we can start at 200 and work our way down we want to do <math>/2</math> the most so if we come across an odd number we just subtract 1. So <math>200/2</math>=<math>100</math>, <math>100/2</math>=<math>50</math>, <math>50/2</math>=<math>25</math>, <math>25-1</math>=<math>24</math>, <math>24/2</math>=<math>12</math>, <math>12/2</math>=<math>6</math>, <math>6/2</math>=<math>3</math>, <math>3-1</math>=<math>2</math>, and <math>2/2</math>=<math>1</math>. We made our way back to 1 but since it is the amount of times the button is pressed than the answer should be <math>\boxed{\textbf{(B)}\ 9}</math>. |
==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 15:33, 28 September 2020
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
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First we can start at 200 and work our way down we want to do 
the most so if we come across an odd number we just subtract 1. So 
=
, 
=
, 
=
, 
=
, 
=
, 
=
, 
=
, 
=
, and 
=
. We made our way back to 1 but since it is the amount of times the button is pressed than the answer should be 
.
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. 
