Difference between revisions of "2005 AIME II Problems/Problem 6"
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== Solution == | == Solution == | ||
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 == | ||
| + | Let the piles before combining them be as follows | ||
| + | |||
| + | <math>A</math> <math>B</math> | ||
| + | <math>1 | ||
| + | 2 | ||
| + | 3 | ||
| + | 4 | ||
| + | . | ||
| + | . | ||
| + | . | ||
| + | n</math> | ||
== See also == | == See also == | ||
Revision as of 01:20, 30 June 2016
Contents
Problem
The cards in a stack of 
cards are numbered consecutively from 1 through from top to bottom. The top 
cards are removed, kept in order, and form pile 
The remaining cards form pile 
The cards are then restacked by taking cards alternately from the tops of pile 
and 
respectively. In this process, card number 
becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles 
and 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.
Solution
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 
numbers must be in front of it. There are 
cards from each of piles A, B in front of card 131. This suggests that 
; the total number of cards is 
.
Solution 2
Let the piles before combining them be as follows
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. 
