Let n_k be the number of painted cells with exactly k painted neighbors, and let N be the number of common sides of painted cells. Each common side belongs to exactly two painted cells, so

Since N is an integer, n₁ + n₃ is even and thus can’t be 25.

(From Prasolov’s Problems in Plane and Solid Geometry.)

If the bath’s volume is V, then the cold tap will fill it at a rate of V/3 and the hot tap at a rate of V/4. It will drain at a rate of V/2. Let t be the time it will take to fill it with the plug out and both taps on. Then

(V/3 + V/4 – V/2)t = V

(1/3 + 1/4 – 1/2)t = 1

and t = 12.

(Via Barry R. Clarke’s Mathematical Conundrums, 2023.)

Well, suppose otherwise. If r and c aren’t equal then one is larger; suppose r > c. Draw a circle around the largest number in each row and a square around the next largest. Each of the circled numbers is at least r/2 (because each is the largest in its row), so each must occupy a different column. Now consider the largest number with a square around it; call that x, and call the circled number in its column y. Let z be the number in y‘s row that has a square around it. Now y + z = r (because they’re the two marked numbers in a row) and y + x = c. Because the largest number with a square around it is x, x must be greater than or equal to z, but this is incompatible with our assumption that r > c. So the assumption is wrong and r and c must be equal.

From MIT mathematician Tanya Khovanova’s blog, via Peter Winkler’s Mathematical Puzzles, 2021.

Effeminacy.

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.)