### Solution to October's Puzzle: Alternating Lamp States

In last month's puzzle, there were 100 lamps arranged in a row that were called lamp 1, lamp 2, lamp 3, and so on. All lamps were originally turned off. Each lamp had a switch that alternates its state (on/off). One hundred people (person 1, person 2, person 3, and so on) were given the following mission: Go to each *n*th lamp and alternate its state, where *n* is the person's number. So person 1 alternated the state of lamps 1, 2, 3, and so on. Person 2 alternated the states of lamps 2, 4, 6, and so on. What's the state of the lamps after all 100 people finished their missions?

The solution to this puzzle is mathematical in its heart. Integers that have an integer square root (e.g., 1, 4, 9, 16) have an odd number of divisors. For example, 9 has the divisors 1, 3, and 9. All other integers have an even number of divisors. For example, 10 has the divisors 1, 2, 5, and 10. The reason for the difference is that an integer with an integer square root can be formulated by multiplying a number by itself, so you don't count the same divisor twice. The formulation of all other numbers is always done in pairs of different divisors: 1*10, 2*5. Thus, all lamps in positions that have an integer square root will end up in the opposite state to the one they started with, while all other lamps will stay in their original state.

### November's Puzzle: Cutting a Stick to Make a Triangle

This month's puzzle is combinatorial. You cut a stick in two random places. What's the probability that you'll be able to form a triangle out of the three pieces?