Difference between revisions of "2005 AMC 8 Problems/Problem 24"
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==Solution== | ==Solution== | ||
− | First we can start at 200 and work our way down to 1. Since we want to | + | First we can start at 200 and work our way down to 1. Since we want to press the button that multiplies by 2 the most, but we are going down instead of up, so we divide by 2 instead. If we come across an odd number, we will subtract that number by 1. |
+ | <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 down to 1 but since it is the amount of times the button is pressed then the answer should be <math>10-1</math>=<math>\boxed{\textbf{(B)}\ 9}</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 16:03, 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 press the button that multiplies by 2 the most, but we are going down instead of up, so we divide by 2 instead. If we come across an odd number, we will subtract that number by 1.
=
,
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,
=
,
=
,
=
,
=
,
=
,
=
,
and
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We made our way down to 1 but since it is the amount of times the button is pressed then 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.