Place the rectangle’s lower left corner at the origin of a checkerboard grid each of whose squares has side 0.5. Now, because each of the smaller rectangles has at least one integer side, each will be colored half white and half black. Consequently the large rectangle itself must be half white and half black.

Now suppose that the large rectangle does not have integer width. What heights will permit it to contain equal areas of black and white? Only those with precisely an even number of checkerboard rows — the integers.

From Stan Wagon, “Fourteen Proofs of a Result about Tiling a Rectangle,” American Mathematical Monthly, Vol. 94 (1987), pp. 601-617, via Peter Winkler, Mathematical Puzzles: A Connoisseur’s Collection (2004).

1. Qc8 Kh6 2. Qh8#. (If 1. … h3 then 2. Qxh3#.)

The chance of rolling no sixes is

5/6 × 5/6 × 5/6 × 5/6 × 5/6

The chance of rolling exactly one six is

1/6 × 5/6 × 5/6 × 5/6 × 5/6

… times 5, because the six can appear on any one of the five dice. So the probabilities are equal.

Put both weights in one pan and pour sugar into the other until it balances. Then remove the weights and pour sugar into the first pan until it balances again. The first pan will contain exactly 10 pounds of sugar.

When successive powers of 5 are divided by 7, the remainders form a repeating series:

5¹ / 7 = 0 remainder 5

5² / 7 = 3 remainder 4

5³ / 7 = 17 remainder 6

5⁴ / 7 = 89 remainder 2

5⁵ / 7 = 446 remainder 3

5⁶ / 7 = 2232 remainder 1

5⁷ / 7 = 11160 remainder 5

5⁸ / 7 = 55803 remainder 4

5⁹ / 7 = 279017 remainder 6

5¹⁰ / 7 = 1395089 remainder 2

…

The 999,999th term of the series is 6.

It’s 1 in 5. Six possible pairings can be drawn from the hat:

red 1 / red 2
red 1 / white
red 2 / white
red 1 / blue
red 2 / blue
white / blue

We know that the last has not been drawn. That leaves 5 possible combinations, so the possibility of red 1 / red 2 is 1 in 5.

“Many people cannot accept that the solution is not 1 in 3,” writes Erwin Brecher, “and of course it would be, if the balls had been drawn out separately and the color of the first ball announced as red before the second had been drawn out. However, as both balls had been drawn together, and then the color of one of the balls announced, then the above solution, 1 in 5, must be the correct one.”

1. Qh7 (K moves) 2. e8=Q#.

Turn over both hourglasses. When the 4-minute glass has run out, turn it over. When the 7-minute glass finishes, turn it over and let it run until the 4-minute glass has run out. At this point we’re at the 8-minute mark and the lower bulb of the 7-minute glass contains 1 minute’s worth of sand. Turn it over to measure the final minute.

It’s better not to spin it. Two of the six chambers are loaded, so my chance of surviving was 4/6. But you know that the hammer fell on an empty chamber. Of the four empty chambers, only one is followed by a loaded one. So if you simply pull the trigger, you have a 3/4 chance of surviving. That’s better than 4/6, so you’re better off not spinning the cylinder.