She’s on the right track. Lori can stop the worms by installing a gutter around the canopy’s perimeter, but one that extends inward, under its lip, rather than outward:

Now a worm that drops from the ceiling onto the canopy and creeps beneath will pass under this gutter and find itself surrounded by a moat, of sorts, that prevents it from reaching the underside of the canopy.

From MIT’s Balint Virag, via Dartmouth mathematician Peter Winkler’s excellent Mathematical Mind-Benders (2007). Winkler adds, “If the worms have no higher-up way to get in her bedroom, Lori can more easily accomplish this same task by encircling the room itself with a water-filled gutter.”

1. e6! and the pawn or bishop will mate.

From Chess Problems, 1873.

Each gear weighs either an even or an odd number of grams. Any set of 12 gears can be divided into two groups of equal weight, so a set of 12 gears has a total weight that’s even. This weight remains even if one of the 12 gears is exchanged with the 13th gear, and this is true no matter which of the 12 is exchanged. So either all the gears are of even weight or all are of odd.

Now find the weight of the lightest gear and subtract it from all 13 gears. This gives us a set of 13 gears that still fulfill the conditions of the problem (though one or more of them now has zero weight). But now we know that each gear has an even weight: If all had originally been even, then we’ve subtracted an even number to produce an even number. And if all were originally odd, then we’ve subtracted odd from odd to produce even.

Now, since each gear has even weight, we can divide the weight of each gear by 2 to obtain another set of 13 gears that still fulfills the conditions of the problem. If all the original gears had the same weight, we can continue to divide by 2 repeatedly without a problem: With each division, we produce a new set that satisfies the conditions, and every gear’s weight will have the same parity, even or odd. But if the original group of gears contained gears of differing weights, then repeated division by 2 will eventually produce a situation in which some weights are even and some odd. No such set can satisfy the original conditions of the problem. So all the original gears must have had the same weight.

Call a door internal if it opens between rooms or external if it opens outside. And designate each room either O (if it has an odd number of internal doors) or E (if even).

Each E room has an even number of external doors, since all rooms have an even number of doors and even – even = even. So the E rooms, taken as a group, have a total number of external doors that’s even.

The total number of O rooms is either odd or even. If it’s odd, then the total number of doors in the O rooms is odd, since the sum of an odd number of odd numbers is odd. But this is impossible: Every internal door must open on two rooms, and this would leave one door that has no second room to open on. So the total number of O rooms is even.

Since each O room has an odd number of internal doors and an even number of total doors, each must have an odd number of external doors. Because the total number of O rooms is even, this means that the O rooms taken as a group have an even number of external doors (since the sum of an even number of odd numbers is even).

Since both groups, the E rooms and the O rooms, have an even number of external doors, so does the whole house.

L.A. Graham provided a simpler solution in 1959.

“The pointing of the words is the solution, as follows: Thou shalt goe thither well and safely, and from thence come home alive againe never: at that field shalt thou be slaine.”

See “Ambiguous Lines.”

Black’s pawns have made 14 captures altogether, which accounts for every one of the missing white pieces. So the missing piece cannot be white. Both kings are in check, which is illegal, so the missing black piece must be on a2, blocking the rook’s check. The missing piece cannot be a queen or rook, as this would put the white king in an illegal double check. So it’s a black knight or a black bishop. And it can’t be a bishop — Black’s light-squared bishop never left its home square (the pawns on b7 and d7 have not moved), and all his pawns are still on the board, so Black has made no promotions. So the missing piece is a black knight on a2.

From Schachmatnaja komposizija, 2007.

If a and b are the first and nth terms of the progression, then the sum of the first n terms is S = (a + b)n/2. So 2S is divisible by n, and since 2S is a power of 2, so is n.

Let x be the number of steps on the escalator. I climb x – 20 steps in 60 seconds, and my wife climbs x – 16 steps in 72 seconds. This means the escalator ascends at a rate of 4 steps every 12 seconds, or 20 steps in 60 seconds. Its length is the sum of these 20 steps and the 20 steps I climb, or 40 steps.

1. Qg8! and the queen or rook will mate.

From Österreichische Lesehalle, June 1881.