Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AMC 8 Problems/Problem 24"

(→‎Solution)
m (→‎Solution 1 (Unrigorous))
 
(17 intermediate revisions by 5 users not shown)
Line 4: Line 4:
 
<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)==
−
hI kIDS i am bob AND I SHALL EAT YOU. NEVETR COME AND TRY TO HURT ME. I AM THE GOD OF HACKING
+
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
−
YOur butt smells
+
<math>200 \div 2 = 100</math>, 
 +
<math>100 \div 2 = 50</math>, 
 +
<math>50 \div 2 = 25</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>. Because we only subtracted <math>1</math> when we had to, this is optimal. ~Roy2020
  
−
 
+
can someone please rigor this
−
First we can start at 200 and work our way down to 1.  Since we want to do divide by two 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>=<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 <math>10-1</math>=<math>\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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems/Problem_24&oldid=220364"