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 to 1. Since we want to do divide by two 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>10-1</math>=<math>\boxed{\textbf{(B)}\ 9}</math>! | + | First we can start at 200 and work our way down to 1. Since we want to do divide by two 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>10-1</math>=<math>\boxed{\textbf{(B)}\ 9}</math>! - <math>\boxed{\textbf{Javapost}}</math> |
| − | - <math>\boxed{\textbf{Javapost}}</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 14:39, 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
First we can start at 200 and work our way down to 1. Since we want to do divide by two the most, so if we come across an odd number we just subtract 1. So 
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, 
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, 
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, 
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, 
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, 
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, 
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, 
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, and 
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. We made our way back to 1 but since it is the amount of times the button is pressed than the answer should be 
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! - 
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. 
