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Difference between revisions of "2000 AIME I Problems/Problem 15"
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== Solution ==
 
== Solution ==
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We try to work backwards. At the end all cards are in a row from left to right in sequential order. We reverse the step: Place the rightmost card on the pile, then place the bottom card on top. Let <math>p_{k}</math> be the position of card 1999 counting from the bottom of the stack when there are <math>2\leq k\leq 2000</math> cards in the stack. We have <math>p_{2}=2</math>. Setting up an recurrence, <math>p_{n+1}=n</math> if <math>p_{n}=1</math> and <math>p_{n+1}=p_{n}-1</math> else. Calculating a few terms we see the pattern <math>p_{3 \cdot 2^k - n}= n+1</math> for <math>n<3 \cdot 2^{k-1}</math>.
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We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, 2(1024 - 1000) = 48, to be exact. Once these cards go through, 1999 will be the <math>512 - 48 = 464th</math> card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position <math>464 \times 2 = 928</math>, meaning that there were <math>927</math> cards on top of it.  
  
 
<math>927</math> cards are above the one labeled <math>1999</math>.
 
<math>927</math> cards are above the one labeled <math>1999</math>.

Revision as of 21:02, 16 March 2009

Problem

A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled 1999?

Solution

We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, 2(1024 - 1000) = 48, to be exact. Once these cards go through, 1999 will be the $512 - 48 = 464th$ card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position $464 \times 2 = 928$, meaning that there were $927$ cards on top of it.

$927$ cards are above the one labeled $1999$.

Template:Incomplete

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
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All AIME Problems and Solutions
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