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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)==
−
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.
+
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
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  <math>200/2</math>=<math>100</math>,   
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  <math>200 \div 2 = 100</math>, 
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  <math>100/2</math>=<math>50</math>,   
+
<math>100 \div 2 = 50</math>,   
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  <math>50/2</math>=<math>25</math>,  <math>25-1</math>=<math>24</math>,
+
  <math>50 \div 2 = 25</math>,
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  <math>24/2</math>=<math>12</math>,
+
<math>25-1 = 24</math>,   
−
<math>12/2</math>=<math>6</math>,
+
  <math>24 \div 2 = 12</math>, 
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<math>6/2</math>=<math>3</math>, <math>3-1</math>=<math>2</math>,  
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<math>12 \div 2 = 6</math>,
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and 
+
  <math>6 \div 2 = 3</math>, 
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<math>2/2</math>=<math>1</math>.  
+
<math>3-1 = 2</math>,  
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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 <math>10-1</math>=<math>\boxed{\textbf{(B)}\ 9}</math>! - <math>\boxed{\textbf{Javapost}}</math>
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  <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"?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution 1 (Unrigorous)

We can start at $200$ and work our way down to $1$. We want to press the button that multiplies by $2$ the most, but since we are going down instead of up, we divide by $2$ instead. If we come across an odd number, then we will subtract that number by $1$. Notice 

$200 \div 2 = 100$,  
$100 \div 2 = 50$,  
$50 \div 2 = 25$,
$25-1 = 24$,  
$24 \div 2 = 12$,  
$12 \div 2 = 6$,  
$6 \div 2 = 3$,  
$3-1 = 2$, 
$2 \div 2 = 1$.

Since we've reached $1$, it's clear that the answer should be $\boxed{\textbf{(B)}\ 9}$- $\boxed{\textbf{Javapost}}$.

Solution 2 (Bounding) - ike.chen

Clearly, there exists a construction for $9$ keystrokes, as shown above. Now, we show this is the smallest possible number of keystrokes.

If there are at most $7$ keystrokes, then the highest number we can reach is $128 < 200$.

If there are $8$ keystrokes, then we consider the following cases:

- $8$ [x2]: This will clearly result in $256$, which isn't desired.

- $7$ [x2], $1$ [+1]: The two largest numbers we can reach from this case are $256$ and $192$, so we know this combination will not work.

- If we use at most $6$ repetitions of [x2], then our number will be at most $(1 + 2) \cdot 2^{6} = 192$, so all of these combinations are bad.

Hence, $\boxed{\textbf{(B)}\ 9}$ 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

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