2005 AIME II Problem 6

Below is the professionally curated solution for Problem 6 of the 2005 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AIME II solutions, or check the answer key.

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Concepts:permutationsparitylinear equation

Difficulty rating: 2560

6.

The cards in a stack of 2n2n cards are numbered consecutively from 11 through 2n2n from top to bottom. The top nn cards are removed, kept in order, and form pile A.A. The remaining cards form pile B.B. The cards are now restacked into a single stack by taking cards alternately from the tops of pile BB and pile A,A, respectively. In this process, card number (n+1)(n + 1) is the bottom card of the new stack, card number 11 is on top of this card, and so on, until piles AA and BB 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 called magical. For example, eight cards form a magical stack because cards number 33 and number 66 retain their original positions. Find the number of cards in the magical stack in which card number 131131 retains its original position.

Solution:

The new stack, read from the bottom up, is n+1, 1, n+2, 2, , 2n, n.n+1,\ 1,\ n+2,\ 2,\ \ldots,\ 2n,\ n. So pile BB's cards occupy the even positions from the top in reverse order, and pile AA's cards occupy the odd positions in reverse order: a card at original position ini \le n (pile AA) moves to position 2(ni)+1,2(n - i) + 1, while a card at position i>ni \gt n (pile BB) moves to position 2(2ni)+2.2(2n - i) + 2.

Since 131131 is odd, card 131131 can keep its position only if it comes from pile A,A, so 131=2(n131)+1,131 = 2(n - 131) + 1, which gives n=196.n = 196. Indeed 131196,131 \le 196, and the stack is magical because card 262262 from pile BB also stays fixed: 2(2n262)+2=262.2(2n - 262) + 2 = 262. The stack has 2n=3922n = 392 cards.

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