Yes. Color the board’s squares like so:

This gives us 22 gray, 21 black, and 21 white squares. We can accomplish our task if each tromino covers squares of three different colors.

Where shall we put the monomino? It can’t go in the bottom left square, for example, as that would leave us with 22 gray, 21 white, and 20 black squares. By symmetry we can also reject the board’s three other corner squares, which might equally well have been colored black, ruining our plan by the same reasoning.

Extending this idea, the monomino can’t go on any white or black square or on any square symmetric to a white or black square. Of the 22 gray squares on the board, only four are symmetric only to other gray squares. If the monomino is placed on one of these squares, the remainder of the board can be covered with 21 trominoes. A sample tiling is shown below.

From J.A.H. Hunter and Joseph S. Madachy, Mathematical Diversions, 1963.

No. This solution is by Adilet Otemissov and the Ashland University Problem Solving Group. If there are m red and 20 – m blue players at the start, then the red player will always be 20 – m times more likely to find a blue bag than to find his own bag on opening the kth locker, no matter the value of k. So the probability that he finds his own bag is

In our case, p(10) = 1/11, so the red player shouldn’t accept — his expected winnings are

which is negative.

(Derek Smith, “The Playground!”, Math Horizons 19:3 [February 2012], 30-33.)

At first this looks absurdly easy — surely promoting either white pawn leads directly to mate?

The trouble is that once either pawn advances, the other is pinned and can’t move. Instead, the key is the crazy-looking 1. Qh7! Black’s rook has no safe refuge on the eighth rank, and if it takes the queen then a pawn promotion is fatal.

From The Problemist, 1941.

07/29/2025 My description isn’t quite accurate — a few readers have pointed out that Black’s rook is safe on b8. In that case White can mate with 2. Nc7, a smothered checkmate. Thanks to everyone who wrote in about this.

Dave: ♠6 ♠7 ♠8 ♠9 ♠10

Frankie: ♣2 ♣3 ♣4 ♣6 ♣7

Georgiana: ♣9 ♣10 ♠J ♣Q ♠K

Cleo: ♣8 ♣J ♣K ♣A ♠A

(By Bob Stanton.)

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.

07/20/2025 UPDATE: From reader Catalin D. Voinescu:

Even simpler, from reader Joe McVeety: “Since you are not given the speed of the river, it must not matter. Therefore, set it to 0. The raft never moves. Boats 1 and 2, traveling at the same speed, will cover the same distance in the same time. When Boat 1 reaches B, Boat 2 is reaching A. The raft never moved, so it is equidistant to Boat 2 and Point A. 0, in this case.”

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.)