**Which Switch? Brainteaser**

There are three on/off switches on the ground floor of a building. Only one operates a single lightbulb on the third floor. The other two switches are not connected to anything. Put the switches in any on/off order you like. Then go to the third floor to see the bulb. Without leaving the third floor, can you figure out which switch is genuine? You get only one try.

*[Answer:] **Turn on the first two switches and turn off the third. Wait 10 minutes and then turn the first switch off and go to the third floor. If the bulb is off but warm, the first switch is the one that operates it; if it is on, the second switch is genuine; and if it is off and cold, the third switch is real.*

**Cryptarithm**

In these problems, each letter corresponds to a single digit. For instance, can you figure out which digit each letter represents to make the sum at the right work?

[**Answer:]**

*S = 3
E = 0
V = 8
N = 7
F = 2
O = 1
R = 5
T = 6
Y = 4*

*To begin solving this problem, notice that 7 times N equals something ending in 9, so N must be 7. From this information, you then know that 7E + 4 equals something that has Y as its units digit and can start experimenting. For instance, E cannot be 1 because then Y would be 1, too, and if E is 2, then V can't be 1. Continuing to reason in this fashion ultimately leads to the answers above.*

**Perplexing Polyhedron **

Can you picture a solid pyramid consisting of a square base and four equilateral triangles, alongside a solid tetrahedron with four faces identical to those of the pyramid's triangles? Now glue one triangle face of the pyramid to a triangle on the tetrahedron. How many faces does the resultant polyhedron have? It's not seven!

*[ Answer:] In the combined solid there are two places at which two triangular faces—one from each constituent solid—perfectly align to form a single face. As a result, the total number of faces for the new polyhedron is only 5 (one square, two triangles and two rhombi).*

**Isogon Proof **

Think about the class of polygons known as serial isogons. All adjacent sides meet at 90 degrees, and the sides are of increasing length: 1, 2, 3, 4, and so on. The simplest isogon, with sides 1–8, is shown at the right. This is the only serial isogon known to tile the plane. But there are more isogons. Can you prove that the number of their sides must always be a multiple of 8?

*[ Answer:] The task is to prove that for any 90-degree serial isogon, the number of sides must be a multiple of 8. *

*Imagine an isogon to be a closed path, traced over the lattice lines of standard graph paper. Assume that the path begins with a unit move to the east; moving east or north is positive and moving south or west is negative. You can describe such a path by placing a plus or minus sign in front of each number in the sequence of moves to indicate the direction of the move. For example, the 8-sided isogon shown in the puzzle sampler has this formula:
+1 + 2 – 3 – 4 – 5 – 6 + 7 + 8.*

*Because the path must close to form an isogon, the sum of all horizontal moves—the odd numbers—must be zero and the sum of all the numbers will also be zero. Also because the path is closed, we know that the number of sides must be a multiple of 4, say 4k. (It is easy to see that it can never be 4 itself—something Martin Gardner joked is a “4-gon conclusion.”) Then the north-south moves are the even-length ones, 2, 4, … 4k. The total north-south distance is therefore 2(1 + 2 + … + 2k) = 2k(2k + 1). Half of this, k(2k + 1), must be north and half of it south.*

*But if k is odd, this distance is odd and cannot be the sum of even-length moves. For instance, consider an isogon of 12 sides (in which case, k = 3). Even numbers in its formula (2, 4, 6, 8, 10, 12) add to 42. If the formula describes a closed path, the sum of the positive numbers in this sequence must equal 42/2 = 21. But no set of even numbers can add to 21. Hence, no formula can be constructed to describe an isogon of 12 sides.*

*This example shows that if the number of sides is a multiple of 4 but not of 8 (all cases in which k is odd, like 12 the last segment of the path, which is vertical, cannot return to the horizontal lattice line that passes through the path’s origin point: the sum of the positive segments going north cannot equal the absolute sum of the negative segments going south. The path end will always end some even number of units above or below the zero horizontal line. *

**Unattacked Queens**

The properties of chess pieces play a part in many challenges, including in a group of problems about unattacked queens. Imagine three white queens and five black queens on a 5 × 5 chessboard. Can you arrange them so that no queen of one color can attack a queen of the other color? There is only one solution, excluding reflections and rotations.

*[ Answer:]
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