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 | + | <math>200 \div 2 = 100</math>, |
− | <math> | + | <math>100 \div 2 = 50</math>, |
− | <math> | + | <math>50 \div 2 = 25</math>, |
− | <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>. | ||
+ | |||
+ | ==Solution 2 (Bounding) - ike.chen== | ||
+ | Clearly, there exists a construction for <math>9</math> keystrokes, as shown above. Now, we show this is the smallest possible number of keystrokes. | ||
+ | |||
+ | If there are at most <math>7</math> keystrokes, then the highest number we can reach is <math>128 < 200</math>. | ||
+ | |||
+ | If there are <math>8</math> keystrokes, then we consider the following cases: | ||
+ | |||
+ | - <math>8</math> [x2]: This will clearly result in <math>256</math>, which isn't desired. | ||
+ | |||
+ | - <math>7</math> [x2], <math>1</math> [+1]: The two largest numbers we can reach from this case are <math>256</math> and <math>192</math>, so we know this combination will not work. | ||
+ | |||
+ | - If we use at most <math>6</math> repetitions of [x2], then our number will be at most <math>(1 + 2) \cdot 2^{6} = 192</math>, so all of these combinations are bad. | ||
+ | |||
+ | Hence, <math>\boxed{\textbf{(B)}\ 9}</math> is the answer. | ||
==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 22:47, 23 October 2021
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
-
.
Solution 2 (Bounding) - ike.chen
Clearly, there exists a construction for keystrokes, as shown above. Now, we show this is the smallest possible number of keystrokes.
If there are at most keystrokes, then the highest number we can reach is
.
If there are keystrokes, then we consider the following cases:
- [x2]: This will clearly result in
, which isn't desired.
- [x2],
[+1]: The two largest numbers we can reach from this case are
and
, so we know this combination will not work.
- If we use at most repetitions of [x2], then our number will be at most
, so all of these combinations are bad.
Hence, is the answer.
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.