Each batter either gets on base or is put out. A player who gets on base either scores, makes an out, or is left on base when the inning ends. Hence each plate appearance will be recorded as an out, a run, or a player left on base:

PA = outs + runs + left on base.

In this game, we know that Casey appeared at the plate four times and was third in the lineup. So Casey, Flynn, and Blake each had four plate appearances, and their six teammates had three apiece. Thus the total number of plate appearances was 30.

There are three outs per inning, so a nine-inning game produces 27 outs. Using the formula above,

30 = 27 + runs + left on base.

Thus the number of runs plus the number of players left on base is 3. But the bases were loaded when Casey came up for the last time, and all three of these players were left on base when he struck out. So Mudville scored no runs in the game.

(Jerry Butters and Jim Henle, “Baseball Retrograde Analysis,” Mathematical Intelligencer 40:4 [December 2018], 71-76.)

From George Arnold et al., The Magician’s Own Book, 1857.

It’s tempting to move the bishop, giving check, and then try to mate with the queen from b7, but the black king can run to f6 and escape. And checking with the queen along the long diagonal allows Black to flee via g8. The answer is the quiet 1. Kh6! and now, no matter which black pawn promotes, White has a ready mate. (If Black runs to g8, then 2. Be6 ends things neatly.)

From The Problemist, 1974.

Three 10 kg weights will get the result we want — dividing this set in two will always leave us with one group that weighs no more than 10 kg. To see that a more massive set must fail, choose any weight in that massive set and add more weights, one by one, until the total mass M in that group exceeds 10 kg. If we are to succeed, the remaining mass m must not exceed 10 kg. But if a is the mass of the last weight we added to the first group, we know that a ≤ 10 kg and that M – a ≤ 10 kg. So the combined mass of all the weights is M + m = (M – a) + a + m ≤ 30 kg.

From the November-December 1994 issue of Quantum.

In order to avoid falling, the lemming must follow some closed loop on the mountaintop. This loop can’t cross itself, because adjacent arrows at the crossing point would disagree by more than 45 degrees. So it’s just a simple loop. Is this possible? There are various ways to show that it’s not; one is to assume that it is possible and then to show that this assumption leads to a contradiction.

If such loops are possible, then they might come in various sizes. One of these must be the smallest possible (meaning that it encloses the smallest number of squares). Suppose the path along this loop runs clockwise. Rotate every arrow on the board 45 degrees clockwise. The resulting board must be legal, because no arrow has rotated relative to its neighbors. But now we seem to have defined a new smallest path — because of the new orientation of the arrows, a lemming dropped inside the ring we’d defined can never escape it (see the diagrams here).

This means that the lemming on this new board must be following some closed loop smaller than the “smallest” one we’d been considering. That’s a contradiction, so our initial assumption must be wrong: No legal board can exist with a closed loop that can save the lemming.

(Via Ravi Vakil, A Mathematical Mosaic, 1996. Peter Winkler discusses this also in Mathematical Puzzles, 2021.)

As you might guess, the dot is back on top.

“Surprisingly, however, the ball makes two complete revolutions in achieving this result,” writes Puzzle Corner columnist Allan Gottlieb. “This can be seen by comparing the described situation with the cycloid generated by the black dot, if the ball were rolling on a flat surface. When the ball is halfway through its travels, the dot touches the fixed surface, which is at the top of the ball for the actual circular surface. Hence, one complete revolution has occurred when the bottom of the circle is reached, and the second revolution occurs in the remaining travel back to the top of the circle.”

The same puzzle has been rolling around in two dimensions since at least 1868.

1. bxa8=B#.

Easy once you see it!

(Bruce Hayden, “The Great Sam,” Chess Review 27:1 [January 1959], 77-78.)

“King Charles the First walked and talked; half an hour after, his head was cut off.”

From Peter Puzzlewell, A Choice Collection of Riddles, Charades, and Rebuses, 1794.

Add a comma after cannot.

Anton was the last passenger on and the second one off, so no one entered the elevator after him and only one person exited before him. He went up at least two floors. Edsel got off before Gerda got on, so Edsel must be the passenger who exited the elevator before Anton did. This means that Filbert entered the car sometime before Anton and left it after him, which means he went up at least four floors. Edsel traveled exactly twice as many floors as Filbert did, so Edsel went up at least eight floors. If we assume that Filbert went up only four floors, then at a minimum we have this:

Floor 1: Edsel gets on
Floor 9: Edsel gets off
Floor 10: Gerda gets on
Floor 11: Filbert gets on
Floor 12: Anton gets on
Floor 13: No one gets on or off
Floor 14: Anton gets off
Floor 15: Filbert gets off
Floor 17: No one gets on or off

This scheme leaves exactly four floors (16, 18, 19, 20) for four passengers (Bessie, Clara, Dagwood, and Gerda) to get off, so this minimum scheme must be correct, and Filbert cannot have risen more than four floors.

So that gives us 11 known floor stops (1, 9, 10, 11, 12, 14, 15, 16, 18, 19, 20). Five of these are odd and six even. We’re told that the elevator’s doors opened at more odd-numbered floors than even-numbered floors, so there were eight stops at odd floors, and the three in addition to the ones we’ve named must be 3, 5, and 7. These must be the floors on which Dagwood, Bessie, and Clara (in that order) got on. Bessie went up more floors than Dagwood, so he must have got off at 16 (after rising 13 floors) and then Bessie got off at 19 or 20 (after rising 14 or 15 floors). If Dagwood had gotten off at floor 18 or higher then Bessie would have had to get off above floor 20. We’re told that no two passengers traveled the same number of floors, so Gerda, who got on at 10, couldn’t have exited at 18. We also know Gerda wasn’t the last one to exit, so she must have got off at 19, which means Bessie got off at 20 and Clara got off at 18:

Anton: on at 12, off at 14, rising 2
Bessie: on at 5, off at 20, rising 15
Clara: on at 7, off at 18, rising 11
Dagwood: on at 3, off at 16, rising 13
Edsel: on at 1, off at 9, rising 8
Filbert: on at 11, off at 15, rising 4
Gerda: on at 10, off at 19, rising 9

From Games magazine, April 1993.