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 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>.   
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>.
+
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
  
==Solution 2 (Bounding) - ike.chen==
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can someone please rigor this
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}}

Latest revision as of 16:57, 1 December 2024

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

can someone please rigor this

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
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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