Solve the logic problem (see below). At the end of the game, the cards must be stacked like so:
7H 8C 5H 5S 5C 3C AS 5D 4C KC 2D AH 7D Joker 3S KS 7S 4H 2H 6H JS 8H 9H TH 7C QH 2C JC 6C 3H 4D AC 6D 2S 8D 9D TD JH QD KH AD 9C 3D 4S QC 6S TC 8S 9S TS JD QS KD
Calculate your score by taking the point value of each card at the top of a stack. These point values — black cards at face value, red cards starting at 14, as suggested — are as follows:
20 8 18 5 5 3 1 18 4 13 15 14 20 wild
This adds up to 144 (plus a bonus), confirming the title of the puzzle. But more importantly, if you take the number of points you get for each stack and read it off as a letter, it spells out THREECARDMONT*, where * represents the wildcard. Therefore the answer to the puzzle is THREE-CARD MONTE.
A method of solving the logic problem follows.
First of all, you know that only one card is moved twice, and there are five moves involving Fives. Since there are only four Fives in the deck, one of these must be the one that's moved twice. All other cards are moved at most once each. So at the end of the game, any card that's not a Five will be either in the stack it started out in, or still in the first stack it was moved to.
Now note that, in the course of the game, only two (distinct) Threes are moved. Since all threes start in the third stack, and move from the bottom when a card is placed on top, this means that two cards must have been placed on top of the third stack over the course of the game, and the two bottom cards from that stack (3C and 3S) must have moved, but the other two cards in that stack (3D and 3H) must still be in the same stack with two cards on top of them. Likewise only one Four has moved, and so on. So to start off, we know the positions of the following cards, simply because they can't have moved since the start of the game.
4H 6H 8H 9H TH QH
3H 4D 6D 8D 9D TD JH QD KH
AD 3D 4S 6S 8S 9S TS JD QS KD
At the beginning of the game, the AH is placed on top of the QH, and that's where it must still be at the end of the game. The second move is the QC. According to the list of moves, a Queen can only be placed on a Five. And at this point, no Fives have been moved, so the only way to place the Queen on a Five is to put it on top of the fifth stack. At the end of the game, the QC must still be in the fifth stack. Since all the Fives have been moved at least once by the end of the game (and cards can only be moved from the bottom of a stack), the QC must be on the bottom of the fifth stack when the game ends.
Likewise, we see from the list of rules that the only card that may be placed on top of a Three is a Seven; on a Four, a Five; on an Eight, a Five; on a Queen, an Ace. So the 3H, 4H, QC, and 8H must be covered by a Seven, a Five, an Ace, and a Five.
5 5 AH
7 4H 6H 8H 9H TH QH
3H 4D A 6D 8D 9D TD JH QD KH
AD 3D 4S QC 6S 8S 9S TS JD QS KD
All right. One of the four Fives - call it "5a" - has been moved twice. How can a card get moved twice? It first gets moved to the top of some stack. Then, over the course of the game, all the cards beneath 5a in that stack must get moved out, to push 5a to the bottom of the stack, where it can be moved out again itself. In other words, 5a was moved first to a stack that eventually had all its original cards moved out of it. Aside from the fifth stack (which 5a could not have been moved to the first time it moved), the only two stacks this could be are the second and the seventh. Keep this in mind. Moreover, since 5a is the only card that is moved twice, and the card 5a is placed on top of in its first move gets moved out from under it, 5a must have been placed on a card which was, itself, in its original stack.
At the end of the game, the AS is on top of the seventh stack, and all four Sevens have been moved out of that stack. This means that AS is at the top of a stack of four cards of which one of the following is true: (a) The bottom card of this stack was placed on top of the 7H, which was subsequently moved out from under it. (b) The bottom card of this stack was placed on top of 5a, which was placed on top of 7H, both of which were subsequently moved out.
Suppose (b) is true. Then the bottom card of the stack must be something that can be placed on a Five. According to the list of moves, this is either a Queen or a Nine; but it can't be a Queen, since each move is made only once and we have already accounted for the move where a Queen is placed on a Five. So the bottom card of this stack is a Nine. These are the possible stacks of four cards based on a Nine allowed by the list of moves:
9 Q 7 2 8 5 5 3 K K 4 4 6 6 6 9 9 9 9 9
None of these has an Ace on top. Therefore (b) can't be true; (a) must be true. Note that this means 5a never went through the seventh stack, so it must have gone through the second stack. (That is, 5a must have originally been placed on 2H.) So at the end of the game, the card on the bottom of the second stack must be a card that can have been placed on top of a Five — namely, a Nine. (By elimination, the 9C.)
So the bottom card in the seventh stack at the end of the game is a card that was placed on the 7H. A Two, a Five, and a Ten may be placed on a Seven. So these are the possible stacks of four cards with those bases (ignoring possible stacks which would involve illicitly repeating a move. In particular, no Sevens can appear in this stack, because the same card would have to have been moved out of this stack and then back in, but only 5a can do that):
4 6 2 J 3 K 9 A 3 5 J 9 9 A A 6 6 5 J J 2 2 5 5 J J 9 9 2 2 2 K K 2 2 2 2 5 5 T T T T T
Only one of these has an Ace on top, so that must be the makeup of the seventh stack at the end of the game. This stack has a Two on a Ten (by elimination, the TC). The only other card that can go on a Ten is a King, so the card on top of the TH must be a King.
5 AS 5 K AH
7 4H 6H J 8H 9H TH QH
3H 4D AC 6D 2 8D 9D TD JH QD KH
AD 9C 3D 4S QC 6S TC 8S 9S TS JD QS KD
A Nine can be covered by a Four or a Six, according to the list of moves. The only thing that can be on top of a Four is a Five. But there's already a Five placed on top of the 4H. Since moves are not repeated, any other Four must be on top of a stack, since there's nothing else to place on any Four. This means that 9C can't be covered by a Four, so 9H must be (by 4C), and so 9C is covered by 6C. Likewise, KH can only be covered by a Two or an Eight, but an Eight can only be covered by a Five, and we know that the top of the thirteenth stack is not a Five. So KH must be covered by a Two.
5 AS 5 4C K AH
7 4H 6H J 8H 9H TH QH 2
6C 3H 4D AC 6D 2 8D 9D TD JH QD KH
AD 9C 3D 4S QC 6S TC 8S 9S TS JD QS KD
If 8C does not cover KH, where is it? The list of moves gives us Eight on a King, so it must cover the fourth King, which we haven't found yet. This means the fourth King can't be on top of any stack. All the bases of stacks are already filled, and the only second place in a stack that's not taken is on top of AD, where a King can't go. So the missing King must be in the third place on some stack. A King can't cover an Ace or Jack, by the list of moves; moreover, it can't cover anything that can cover an Ace (that is, a Two, Five, or Jack). So the missing King must cover the 6C. This means the 6H isn't covered by a King, since moves aren't repeated, so it must be covered by a Three.
8C 5 3 AS 5 4C K AH K 7 4H 6H J 8H 9H TH QH 2 6C 3H 4D AC 6D 2 8D 9D TD JH QD KH AD 9C 3D 4S QC 6S TC 8S 9S TS JD QS KD
We have a Two covered by a Jack in the seventh stack. A Five can cover a Two, but we know that the thirteenth stack can't have a Five on top. The only other card that can cover a Two is a Seven, to the thirteenth stack has a Seven on top.
Early in the game, when 5a was placed in the second stack, it covered the 2H. Our list of moves tells us that there must be some other Five placed on a Two sometime; and since 5a is the only card moved twice, there must be a Five still on a Two at the end of the game. So there must be some other Two that's not on the top of a stack, to be covered by the Five. It can't be the eleventh stack, because a Two can't cover a Jack.
Suppose there were a Two covering the AD, and a Five covering the Two. What would go on top of that Five? The list of moves only allows a Nine and a Queen to be placed on a Five, and those moves have already been used to get the 9C and the QC into the second and fifth stacks. So the Two is not on top of the AD. Therefore it must be on the AC, with a Five on top of that. (Note that this must be 5a, since the only way for a card to return to the stack it started in is by moving twice.)
The only card apart from Two that can cover an Ace is a Jack. So there's a Jack on top of AD.
8C 5 5a 3 AS 5 4C K AH 7 K 7 4H 2 6H J 8H 9H TH QH 2 J 6C 3H 4D AC 6D 2 8D 9D TD JH QD KH AD 9C 3D 4S QC 6S TC 8S 9S TS JD QS KD
The Jack in the first stack can be covered by a Seven or a Three. Supose it's a Seven. That Seven can't be covered by a Five because the first stack doesn't have a Five on top at the end of the game; and the Ten in the seventh stack was put on a Seven earlier. From the list of moves, that leaves only a Two to go on top of the first stack. However, the card on top of the first stack is the last card played in the game. Since only 5a is moved twice, this Two must have come direct from the bottom of the second stack. But this would imply that the 9C had been played on top of a Two, which it can't have been — 9C covered 5a. This means the first stack does not have a Two on top at the end of the game, which means the Jack in the first stack can't be covered by a Seven. Therefore the Jack in the first stack must be covered by a Three, which must be covered by a Seven since that's the only card that can cover a Three.
This means that it's the JH in the eleventh stack that must be covered by a Seven.
7 8C 5 5a 3 AS 5 4C K AH 7 3 K 7 4H 2 6H J 8H 9H TH 7 QH 2 J 6C 3H 4D AC 6D 2 8D 9D TD JH QD KH AD 9C 3D 4S QC 6S TC 8S 9S TS JD QS KD
That leaves the third and fifth stack, each of which has a Seven in third place than can be covered by a Two or a Five. Suppose the Two is on top of the third stack and the Five on top of the eleventh stack; let's play the last few moves of the game backward.
The last move was placing the Seven on the first stack. Since only the 5a is moved twice, it came from the bottom of the seventh stack, so the previous move was placing the AS on the seventh stack. Before that the Three was placed on the first stack; before that was the Two, by hypothesis, placed on the third stack. This Two must have come from the bottom of the second stack; but as above, this would imply that the 9C had been placed on top of the Two, which is not correct. That means that the hypothesis is wrong, and at the end of the game the third stack must have a Five on top and the eleventh stack a Two.
7 8C 5 5 5a 3 AS 5 4C K 2 AH 7 3 K 7 4H 2 6H J 8H 9H TH 7 QH 2 J 6C 3H 4D AC 6D 2 8D 9D TD JH QD KH AD 9C 3D 4S QC 6S TC 8S 9S TS JD QS KD
By playing the game backwards, we can find out the suit of each card that we don't yet know, since the order in which cards are played out of the bottom of their original stacks is sufficient to determine their suits. We can reconstruct every move backwards because we know that every move must begin by taking a card from the bottom of the stack it started in, except for the second move of 5a, which was taken from the second stack.
7H 8C 5H 5S 5C 3C AS 5D 4C KC 2D AH 7D 3S KS 7S 4H 2H 6H JS 8H 9H TH 7C QH 2C JC 6C 3H 4D AC 6D 2S 8D 9D TD JH QD KH AD 9C 3D 4S QC 6S TC 8S 9S TS JD QS KD
2006 MIT Mystery Hunt