Yes.

First, note that two consecutive cubes are always separated by one more than a multiple of 6. To see this, note that (n+1)³ = n³ + 3n² + 3n + 1, so (n+1)³ – n³ = 3n(n+1) + 1. n must be either even or odd; if it is even, then n(n + 1) is even, and if it is odd, then n+1 is even, so again, n(n+1) is even. Therefore, 3n(n+1) is divisible by both 2 and 3, so it must be a multiple of 6, so the difference (n+1)³ – n³ is one more than a multiple of 6.

Now, it’s easy to check that Kandice will safely jump over the first two guards (at 1 and 8). Suppose Kandice later gets caught and Gary the Guard is the first guard who catches her. Since passing the guard previous to Gary, Kandice would have finished one jump of size 7 and several jumps of size 6.

Because the distance between the guards is one more than a multiple of 6 (as shown in this illustration), counting backwards we see this pattern would have forced Kandice to land on the previous guard also. But this is impossible, since Gary was the first guard to catch Kandice. So Gary does not exist and Kandice will never be caught!

Key: Cube numbers are the positions of the guards.

Evens and odds alternate, so if some number n isn’t even, then the next number (n+1) must be even.

This intuitive solution is by V. Dubrovsky. Imagine a third point, C, that’s as far upstream from A as A is from B (so CA = AB). And imagine a third motorboat that parallels the movements of the first, setting out from point C and heading downstream at the same time. The third and second motorboats are equidistant from the raft when they begin their journeys, and both move at the same speed relative to the raft (and to the surface of the water). So they’ll remain equidistant from the raft as they progress toward it. When the first motorboat reaches B, the third will reach A. At that moment the raft will be the same distance from the second motorboat and from Point A (now occupied by the third motorboat). That’s the answer.

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.