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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==
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==Solution 1 (Unrigorous)==
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hI kIDS i am bob AND I SHALL EAT YOU. NEVETR COME AND TRY TO HURT ME. I AM THE GOD OF HACKING
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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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YOur butt smells
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<math>200 \div 2 = 100</math>, 
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<math>100 \div 2 = 50</math>, 
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<math>50 \div 2 = 25</math>,
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<math>25-1 = 24</math>, 
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<math>24 \div 2 = 12</math>, 
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<math>12 \div 2 = 6</math>, 
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<math>6 \div 2 = 3</math>, 
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<math>3-1 = 2</math>,
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<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>. Because we only subtracted <math>1</math> when we had to, this is optimal. ~Roy2020
  
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can someone please rigor this
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First we can start at 200 and work our way down we want to do <math>/2</math> the most so if we come across an odd number we just subtract 1.  So <math>200/2</math>=<math>100</math>, <math>100/2</math>=<math>50</math>, <math>50/2</math>=<math>25</math>, <math>25-1</math>=<math>24</math>, <math>24/2</math>=12<math>, </math>12/2<math>=</math>6<math>, </math>6/2<math>=</math>3<math>, </math>3-1<math>=</math>2<math>, and </math>2/2<math>=</math>1<math>. We made our way back to 1 but since it is the amount of times the button is pressed than the answer should be \boxed{\textbf{(B)}\ 9}</math>.
 
  
 
==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 01:39, 11 June 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 (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 $\boxed{\textbf{(B)}\ 9}$- $\boxed{\textbf{Javapost}}$. Because we only subtracted $1$ when we had to, this is optimal. ~Roy2020

can someone please rigor this

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. AMC logo.png

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