Difference between revisions of "2000 AIME I Problems/Problem 15"
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== Solution == | == 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 <math>512 - 48 = | + | 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 = 464^\text{th}</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>\boxed{927}</math> cards are above the one labeled <math>1999</math>. |
== See also == | == See also == | ||
{{AIME box|year=2000|n=I|num-b=14|after=Last Question}} | {{AIME box|year=2000|n=I|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:55, 24 December 2018
Problem
A stack of cards is labelled with the integers from to 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: In the original stack of cards, how many cards were above the card labeled ?
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 card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position , meaning that there were cards are above the one labeled .
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.