Difference between revisions of "2005 AIME II Problems/Problem 6"

(Solution 2)
(Solution 2)
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<math>A-</math>            <math>1, 2, 3, 4, 5 ... n</math>
 
<math>A-</math>            <math>1, 2, 3, 4, 5 ... n</math>
 +
 
<math>B-</math>            <math>(n+1), (n+2), (n+3)...(2n)</math>
 
<math>B-</math>            <math>(n+1), (n+2), (n+3)...(2n)</math>
  

Revision as of 02:22, 30 June 2016

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. 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 $131 - 1 = 130$ 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

Let the piles before combining them be as follows

$A-$ $1, 2, 3, 4, 5 ... n$

$B-$ $(n+1), (n+2), (n+3)...(2n)$

See also

2005 AIME II (ProblemsAnswer KeyResources)
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

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