We're back this week with another logic puzzle. For this one, you'll need all the help you can get.

## Sunday Puzzle #4: The Logician's Children

*Two former college roommates, both logicians, meet at a conference after many years without contact. While catching up, the two eventually get around to discussing their children. The first logician asks the second how many children he has, and what their ages are. The second replies that he has 3 children, but (ever the logician) he will only reveal clues about their ages. The first logician must deduce for himself the ages of the second's children.*

*"First," says the logician, "the product of my children's ages is 36."*

*"Second, the sum of their ages is the same as our apartment number in college."*

*"Third, my oldest child has red hair."*

*Upon hearing the third clue, the first logician replies at once with the ages of his friend's children. What are they? How do you know?*

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Many thanks to BJ Myers for suggesting this week's puzzle. We'll be back next week with the solution – and a new puzzle! Got a great brainteaser, original or otherwise, that you'd like to see featured? E-mail me with your recommendations. (Be sure to include "Sunday Puzzle" in the subject line.)

*Art by Tara Jacoby*

## SOLUTION to Sunday Puzzle #3: The Bear Hunter (Part 2)

Last week, I asked you to think of points *besides the North Pole* from which someone could walk a mile south, a mile east, and a mile north and find herself back at her point of origin.

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To do this, one must travel to the very opposite side of the earth – or rather, just shy of it. Many of you came up with the correct solution (or close to it) in the comments, but an exchange between Peter Ravn Rasmussen and whoa, I believe, was the first to provide a complete answer. Rasmussen wrote:

Any point that is 1 + 1/(2 π) miles north of the South Pole will satisfy the conditions of this riddle's setup.

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The reasoning is as follows: There is a line of latitude near the South Pole with a circumference of one mile. If one begins at a point 1+1/(2 π) miles north of the South Pole and walks one mile south, she will find herself on the mile-round line of latitude; walking one mile east will therefore bring her in a complete circle around the pole, such that when she turns and walks a mile north, she finds herself right back where she started.

But wait! As whoa points out, this answer is *almost* right, but not quite:

To [travel a mile south and find one's self on] a circle of circumference 1/n miles, whose radius r is 1/(2πn), one must travel a portion of the Earth's circumference (call the circumference 2πR) given by the angle one will reach off the Earth's axis. This angle is arcsin(r/R), and so the distance to any point on the circle from the South Pole is R(arcsin(1/2πnR)). That's very close to 1/(2πn) for all values of n.

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The upshot is that one must begin from any point on a circle drawn at a distance slightly more than 1+1/(2π) miles from the South Pole. Alternatively, one could start at various distances closer to the pole, such that one's walk east brings her not once around the Pole, but two times, three times, four times, ..., or n times.

In other words, there are truly an infinite number of starting points to consider!

## Previous Weeks' Puzzles

- "The Hardest Logic Puzzle In The World"
- 100 Green-Eyed Dragons
- Can you figure our this parking lot's numbering system?

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