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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>
  
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==Solution 1 (Unrigorous)==
+
==Solution 1 (Now rigorous)==
 
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
 
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>200 \div 2 = 100</math>,   
Line 15: Line 15:
 
  <math>3-1 = 2</math>,  
 
  <math>3-1 = 2</math>,  
 
  <math>2 \div 2 = 1</math>.   
 
  <math>2 \div 2 = 1</math>.   
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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>.
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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
  
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==Solution 2 (Bounding) - ike.chen==
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==Solution 2 (Binary)==
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Clearly, there exists a construction for <math>9</math> keystrokes, as shown above. Now, we show this is the smallest possible number of keystrokes.
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We make two key observations. First, pressing [x2] appends a <math>0</math> to the end of a number's binary representation. Second, pressing [x2] and then pressing [+1] appends <math>1</math> to the end of a number's binary representation.
  
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If there are at most <math>7</math> keystrokes, then the highest number we can reach is <math>128 < 200</math>.
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The base-ten number <math>200</math> is represented as <math>11001000_2</math> in binary. Therefore, the five <math>0</math>s contribute <math>5</math> button presses. Similarly, each of the three <math>1</math>s each contribute <math>2</math> button presses, although we do not count one of them as the calculator initially starts with the number <math>1.</math> Thus, the answer is <math>5 + 2 \cdot 2 = \boxed{\textbf{(B)}\ 9}</math>
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If there are <math>8</math> keystrokes, then we consider the following cases:
 
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- <math>8</math> [x2]: This will clearly result in <math>256</math>, which isn't desired.
 
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- <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.
 
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- 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.
 
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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}}

Latest revision as of 22:34, 13 May 2025

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 (Now rigorous)

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 $\boxed{\textbf{(B)}\ 9}$- $\boxed{\textbf{Javapost}}$. Because we only subtracted $1$ when we had to, this is optimal. ~Roy2020

Solution 2 (Binary)

We make two key observations. First, pressing [x2] appends a $0$ to the end of a number's binary representation. Second, pressing [x2] and then pressing [+1] appends $1$ to the end of a number's binary representation.

The base-ten number $200$ is represented as $11001000_2$ in binary. Therefore, the five $0$s contribute $5$ button presses. Similarly, each of the three $1$s each contribute $2$ button presses, although we do not count one of them as the calculator initially starts with the number $1.$ Thus, the answer is $5 + 2 \cdot 2 = \boxed{\textbf{(B)}\ 9}$

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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