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: Similar to Solution 5, we treat each run-through of the deck from the lowest-indexed card to the highest-indexed card as a separate round. Notice that after each round, approximately half the deck will remain, with the other half having been cast aside to the right in sorted order. Then, we can model each round as if we are running a "Sieve" through the deck with powers of 2s, filtering out the cards to put to the right and the cards to be pushed to the bottom.
At the beginning, when all $2000$ cards are still in the deck, notice that the first run-through of the deck consists of removing all the cards such that their index $I\equiv 1\pmod{2}$ , while we leave behind all the cards such that their index $I\equiv 2\pmod{2}$ . (Since the cards are not $0$ -indexed, using $2$ instead of $0\pmod{2}$ is simpler to keep track of the cards. This will appear again later.) The remaining $1000$ cards, then, will all share the attribute that their index numbers are even. To split this further for round 2, we look more specifically at each index, categorizing them further as $I\equiv 2\pmod 4$ or $I\equiv 4\pmod 4$ in that order (The card with the smaller remainder will always be the card on the top of the new deck). Then, in Round 2, we will remove all the cards with index values that leave a remainder of $2$ when divided by $4$ , while leaving the multiples of $4$ behind. Notice that each time, in each new deck, we are removing all the cards with an odd position in that specific deck, and leaving behind all the cards with an even position in that specific deck.
This process will continue until we reach  Round 5, where the number of cards that remain ( $125$ ) is no longer a multiple of 2. Therefore, when we remove all the cards with indices $I\equiv 16\pmod{32}$ and leave behind all the cards with indices $I\equiv 32\pmod{32}$ , the last card in the deck will be removed instead of being placed to the bottom, with $125-63=62$ cards remaining. Because of this, when Round 6 begins, we cannot follow our normal procedure, since the top card of the Round 6 deck would have been moved to the bottom! Therefore, when we are using our sieve with $I\equiv 32\pmod{64}$ and $I\equiv 64\pmod{64}$ , the order at which the remainders are presented have flipped, so all the cards with indices that are multiples of 64 will be removed, while all the cards with indices that are 32 more than a multiple of 64 will remain. The $62$ cards from Round 6 will be reduced to $31$ , and the last card in the deck will be removed just like in Round 5, which means Round 7 will be affected the same way. Of the choices of $I\equiv 32\pmod{128}$ and $I\equiv 32+64=96\pmod{128}$ , we will remove the latter, leaving behind the cards with indices that are 32 more than a multiple of 128. However, since 31 is an odd number, the last card of the deck will NOT be removed, which means our sieving process will return to normal after Round 7, with $31-15=16$ cards remaining. Since $16$ is a perfect power of $2$ , we will not need to worry about this scenario again in the following rounds.
Round 8: $I\equiv 32\pmod{256}$ removed; $I\equiv 32+128=160\pmod{256}$ remains; $16-8=8$ cards left.
Round 9: $I\equiv 160\pmod{512}$ removed; $I\equiv 160+256=416\pmod{512}$ remains; $8-4=4$ cards left.
Round 10: $I\equiv 416\pmod{1024}$ removed; $I\equiv 416+512=928\pmod{1024}$ remains; $4-2=2$ cards left.
Round 11: $I\equiv 928\pmod{2048}$ removed; $I\equiv 928+1024=1952\pmod{2048}$ remains; $2-1=1$ card left.
Round 12: The last card in position $1952$ is sorted into place (This is card # $2000$ !).
It is obvious that the single card put into place in Round 11 is the second-to-last card in the deck, which is our target # $1999$ . Then, its position in the original deck leaves a remainder of $928$ when divided by $2056$ . It is easy to see that $928$ is the only index that satisfies this, so there will be $928-1=\boxed{927}$ cards above card # $1999$

Answer:  927