-、(Full marks 10 points) There are two decks of playing cards, each deck arranged in such a way that the first two cards are the Big Joker and the Small Joker, followed by the four suits of Spades, Hearts, Diamonds, and Clubs, with each suit arranged in the order of $1,2,3$, $\cdots, J, Q, K$. Someone stacks the two decks of cards in the above order, then discards the first card, places the second card at the bottom, discards the third card, places the fourth card at the bottom, …, and continues this process until only one card remains. What is the last remaining card?
Ten, if there are only 64 cards in hand, and you discard according to this rule, then the last remaining card will be the 64th card.

Now, if you have 108 cards in hand, which is $108-64=44$ (cards) more, we only need to discard 44 cards according to the rule, and place the 88th card at the bottom of the remaining cards, at which point the number of cards in hand will be 64. This way, the card left after discarding will be the 88th card in the original order, which is: $88-54-2$ $26=6$. According to the suit arrangement, 6 should be the 6 of Diamonds. Therefore, the last remaining card is the 6 of Diamonds.