Difference between revisions of "2005 AMC 8 Problems/Problem 24"
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<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12</math> | <math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12</math> | ||
| − | ==Solution== | + | ==Solution 1 (Unrigorous)== |
| − | + | We can start at <math>200</math> and work our way down to <math>1</math>. We want to press the button that multiplies by <math>2</math> the most, but since we are going down instead of up, we divide by <math>2</math> instead. If we come across an odd number, then we will subtract that number by <math>1</math>. Notice | |
| + | <math>200 \div 2 = 100</math>, | ||
| + | <math>100 \div 2 = 50</math>, | ||
| + | <math>50 \div 2 = 25</math>, | ||
| + | <math>25-1 = 24</math>, | ||
| + | <math>24 \div 2 = 12</math>, | ||
| + | <math>12 \div 2 = 6</math>, | ||
| + | <math>6 \div 2 = 3</math>, | ||
| + | <math>3-1 = 2</math>, | ||
| + | <math>2 \div 2 = 1</math>. | ||
| + | Since we've reached <math>1</math>, it's clear that the answer should be <math>\boxed{\textbf{(B)}\ 9}</math>- <math>\boxed{\textbf{Javapost}}</math>. Because we only subtracted <math>1</math> when we had to, this is optimal. ~Roy2020 | ||
==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}} | ||
Latest revision as of 18:22, 21 December 2023
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 1 (Unrigorous)
We can start at 
and work our way down to 
. We want to press the button that multiplies by 
the most, but since we are going down instead of up, we divide by instead. If we come across an odd number, then we will subtract that number by . Notice
,
,
,
,
,
,
,
,
.
,
,
,
,
,
,
,
,
.
Since we've reached , it's clear that the answer should be 
- 
. Because we only subtracted when we had to, this is optimal. ~Roy2020
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. 
