Eight. The board’s perimeter consists of 14 black and 14 white squares, and one bishop can’t control more than 4 squares on the border. That means we’ll need at least 4 white bishops to cover the 14 white border squares, because 3 bishops could cover at most 12. Similarly we’ll need at least 4 black bishops to cover the 14 black border squares. As it happens, a row of eight bishops across either of the board’s two middle ranks does the job, so the minimum needed is 8.

Imagine the very worst case: You might pack six non-matching shoes, two non-matching socks, six left-hand gloves of one color, and six right-hand gloves of the other. You can make each of these failures impossible by adding one more item. So you’ll need to pack seven shoes, three socks, and 13 gloves.

From Peter Winkler’s excellent collection Mathematical Puzzles, 2021.

04/20/2021 UPDATE: I made a mistake here — in Winkler’s original, you’re not able to distinguish right-handed gloves from left in the dark. So instead of “six left-hand gloves of one color, and six right-hand gloves of the other,” I should have said, “six brown all left or all right, and six tan, also all left or all right,” as in Winkler’s original. In that case 13 gloves are needed.

(Thanks, Chris and Adam.)

1. Nd2!

(Via S.S. Blackburne, Terms and Themes of Chess Problems, 1907.)

Bob is more likely to win.

Label a carton with a “b” (respectively, a “c”) if Bob (respectively, Chris) reaches that carton more quickly, and also record Bob’s (respectively, Chris’s) score upon reaching that carton. Label cartons A and L with “xx” since both players reach those cartons simultaneously.

We obtain:

xx b2 b3 b4
c2 c5 b7 b8
c3 c6 c9 xx

Note that there are five b cartons and five c cartons. So the cases in which Alice selects carton A or carton L are equally split between Bob and Chris. Similarly if Alice selects two b cartons then Bob necessarily wins, but these are balanced out by an equal number of cases in which Alice selects two c cartons and C necessarily wins.

The crucial cases occur when Alice selects one b carton and one c carton. Bob wins if the b carton has a lower score than the c carton:

b2 and (c3 or c5 or c6 or c9)
b3 and (c5 or c6 or c9)
b4 and (c5 or c6 or c9)
b7 and c9
b8 and c9

Chris wins if the c carton has a lower score than the b carton:

c2 and (b3 or b4 or b7 or b8)
c3 and (b4 or b7 or b8)
c5 and (b7 or b8)
c6 and (b7 or b8)

Since this yields 12 cases in Bob’s favor and only 11 cases in Chris’s favor, Bob has the advantage.

No. When colored like a checkerboard, the rectangle has an equal number of light and dark squares. So does each of the tetrominoes … with one exception. So the task is impossible.

1. Bg5! leads to two pretty mates:

1. … Kf5 2. Rc5#
2. … Kd6 2. Bf4#

No one in this country can ever claim to be a hunter — a noble who did so would be lying, and a hunter who did so would be telling the truth. This means that Ali must have been lying when he attributed that claim to Ahmed — which tells us that Ali is a hunter. And if Ali always lies, then his statement that Azab is a hunter is false, which tells us that Azab is a noble. That means that we can trust Azab when he tells us that Ahmed is a noble.

(From Goodman’s Problems and Projects, 1972.)

The prisoners agree that one of them (say, Alice) will try repeatedly to turn the light on, and each of the others will turn it off twice.

That is, if Alice finds the light off she’ll turn it on, and if she finds it on she’ll leave it on. Each of the others will turn it off the first two times he finds it on and otherwise leave its state unchanged.

Alice counts the number of times she finds the room dark after her first visit. When she’s returned to find the room dark 2n – 3 times, she can announce that everyone has visited the room.

Each time Alice returns to the room and finds it dark, that’s a sign that one of the other n – 1 prisoners has visited it. If a prisoner exists who still hasn’t visited the room, then the light can’t have been turned off more than 2(n – 2) = 2n – 4 times. So once Alice’s total exceeds that, reaching 2n – 3, she knows that everyone’s visited the room.

(Note that the light’s initial state doesn’t matter: If the first visitor isn’t Alice and he finds the light on, he’ll turn it off, and that sign of his presence won’t be evident to her. But she’ll have to turn the light on again before he can perform his second off-turning, and that ensures that she’ll register his second visit.)

(From Peter Winkler’s excellent Mathematical Puzzles: A Connoisseur’s Collection, 2003.)