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 AIME II Problems/Problem 6"

(→‎Solution)
 
(6 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
The cards in a stack of <math> 2n </math> cards are numbered [[consecutive]]ly from 1 through <math> 2n </math> from top to bottom. The top <math> n </math> cards are removed, kept in order, and form pile <math> A. </math> The remaining cards form pile <math> B. </math> The cards are then restacked by taking cards alternately from the tops of pile <math> B </math> and <math> A, </math> respectively. In this process, card number <math> (n+1) </math> becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles <math> A </math> and <math> B </math> are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. Find the number of cards in the magical stack in which card number 131 retains its original position.
+
The cards in a stack of <math> 2n </math> cards are numbered [[consecutive]]ly from 1 through <math> 2n </math> from top to bottom. The top <math> n </math> cards are removed, kept in order, and form pile <math> A. </math> The remaining cards form pile <math> B. </math> The cards are then restacked by taking cards alternately from the tops of pile <math> B </math> and <math> A, </math> respectively. In this process, card number <math> (n+1) </math> becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles <math> A </math> and <math> B </math> are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. For example, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical stack in which card number 131 retains its original position.
  
−
== Solution ==
+
== Solution 1 ==
 
Since a card from B is placed on the bottom of the new stack, notice that cards from pile B will be marked as an even number in the new pile, while cards from pile A will be marked as odd in the new pile. Since 131 is odd and retains its original position in the stack, it must be in pile A. Also to retain its original position, exactly <math>131 - 1 = 130</math> numbers must be in front of it. There are <math>\frac{130}{2} = 65</math> cards from each of piles A, B in front of card 131. This suggests that <math>n = 131 + 65 = 196</math>; the total number of cards is <math>196 \cdot 2 = \boxed{392}</math>.
 
Since a card from B is placed on the bottom of the new stack, notice that cards from pile B will be marked as an even number in the new pile, while cards from pile A will be marked as odd in the new pile. Since 131 is odd and retains its original position in the stack, it must be in pile A. Also to retain its original position, exactly <math>131 - 1 = 130</math> numbers must be in front of it. There are <math>\frac{130}{2} = 65</math> cards from each of piles A, B in front of card 131. This suggests that <math>n = 131 + 65 = 196</math>; the total number of cards is <math>196 \cdot 2 = \boxed{392}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
−
Let the piles before combining them be as follows
+
If you index the final stack <math>1,2,\dots,2n</math>, you notice that pile A resides only in the odd indices and has maintained its original order aside from flipping over. The same has happened to pile B except replace odd with even. Thus, if 131 is still at index 131, an odd number, then 131 must be from pile A. The numbers in pile A are the consecutive integers <math>1,2,\dots, n</math>. This all leads us to the following equation.
−
 
+
<cmath>
−
<math>A</math>               <math>B</math>
+
131=2n-2(131)+1\implies2n=\boxed{392}
−
<math>1
+
</cmath>
−
2
 
−
3
 
−
4
 
−
.
 
−
.
 
−
.
 
−
n</math>
 
−
 
 
 
== See also ==
 
== See also ==
 
{{AIME  box|year=2005|n=II|num-b=5|num-a=7}}
 
{{AIME  box|year=2005|n=II|num-b=5|num-a=7}}

Latest revision as of 23:31, 25 July 2018

Problem

The cards in a stack of $2n$ cards are numbered consecutively from 1 through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A,$ respectively. In this process, card number $(n+1)$ becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles $A$ and $B$ are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. For example, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical stack in which card number 131 retains its original position.

Solution 1

Since a card from B is placed on the bottom of the new stack, notice that cards from pile B will be marked as an even number in the new pile, while cards from pile A will be marked as odd in the new pile. Since 131 is odd and retains its original position in the stack, it must be in pile A. Also to retain its original position, exactly $131 - 1 = 130$ numbers must be in front of it. There are $\frac{130}{2} = 65$ cards from each of piles A, B in front of card 131. This suggests that $n = 131 + 65 = 196$; the total number of cards is $196 \cdot 2 = \boxed{392}$.

Solution 2

If you index the final stack $1,2,\dots,2n$, you notice that pile A resides only in the odd indices and has maintained its original order aside from flipping over. The same has happened to pile B except replace odd with even. Thus, if 131 is still at index 131, an odd number, then 131 must be from pile A. The numbers in pile A are the consecutive integers $1,2,\dots, n$. This all leads us to the following equation. \[131=2n-2(131)+1\implies2n=\boxed{392}\]

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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_AIME_II_Problems/Problem_6&oldid=96501"