There’s an obvious answer to this classic puzzle, and there’s the correct answer. Which will you choose?

This week’s puzzle was suggested by reader Eric Ennis. It’s a classic brain teaser that reads a little like a distance problem (i.e. “if Train A leaves the station at such and such a time traveling such and such speed...”), but the solution can be arrived at with zero math whatsoever, through logical, conceptual thinking.

## Sunday Puzzle #30: Blowin’ in the Wind

*An airplane flies in a straight line from airport A to airport B, then back in a straight line from B to A. It travels with a constant engine speed and there is no wind. Will its travel time for the same round trip be greater, less, or the same if, throughout both flights, at the same engine speed, a constant wind blows from A to B?*

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.)

*$13 for 48 AA, $12 for 48 AAA, $8 for 20 AAA, $8 for four 9V batteries

*Image by Jim Cooke*

## SOLUTION To Sunday Puzzle #29: Counting Coins

Last week, I asked you to devise a strategy for sorting fifty coins into two groups,* *such that each group contains the same number of coins oriented heads-up. You were told on the front end that sixteen of the fifty coins were already in the heads-up position. You were also given permission to flip as many or as few coins as you saw fit. The catch? You were also blindfolded.

At first glance, this puzzle seems impossible. But with a little algebra, a surprisingly clear strategy emerges. The first person to explain the algebraic solution was commenter bewareofgeek. You can read their explanation (and the ensuing discussion, which is also worth perusing) here. For the math phobic, commenter Improbable was the first to provide a clearly stated conceptual explanation. It reads as follows:

Put all the coins in one pile, then split off sixteen, flipping each one over as you put it in the second pile. If it was tails, you now have a heads coin to match one of the remaining heads coins. If it was heads, it’s now tails, reducing the number of heads in the first pile but not increasing it in the second.

Whether you solve this puzzle conceptually or algebraically, a general strategy emerges by which k coins, n of which are heads up, can be manipulated and divided into two groups with an equal number of heads-up coins: Simply separate n coins from the original set of k coins, and turn them all over! This is the “one simple rule” by which any formulation of this puzzle can be solved.

So. Imagine that only ten of the fifty coins are oriented heads up, instead of sixteen. Following our rule, we separate ten coins from the group of fifty and turn all ten of them over. We can’t say how many heads each pile will contain, but we *do* know that the number of heads in the pile of ten coins will equal the number of heads in the pile of forty. We can apply the same strategy whether the number of heads-up coins is twenty-seven out of fifty, or three out of one hundred, or five out of 1,000,000. Simply create a second pile of coins equal in number to the number of heads in the entire set, and turn them all over.

When Sunday Puzzle reader Mikko P. first suggested this puzzle, he posed it several ways. Every variation could be solved by the methods described above, but some of them were more challenging than others. Below, I have included what I think is the trickiest formulation of the riddle:

Zamzam is blindfolded and brought to a table by a man dressed in black. The man says to her: “On the table there are a vast number of coins, all of them tails up.”

Zamzam, being in no position to argue, nods.

The man continues: “What is your favourite number?”

Puzzled, Zamzam answers. The man begins flipping the coins over. When he’s finished, he tells her: “Now there are exactly your favourite number of coins heads up. Your task is to separate the coins into two groups, flipping coins as you see fit, so that each group has the same amount of coins that are heads up.”

The man puts gloves on Zamzam’s hands so that she can’t feel the orientation of the coins, and says: “Begin.”

How is Zamzam able to solve the problem?

To someone unfamiliar with the solution, the uncertainty surrounding the total number of coins (which is simply described as “vast”) and the lack of information vis-à-vis Zamzam’s favorite number can make this already difficult problem seem that much more challenging. In fact, these ambiguities are clues; the puzzle’s vagueness speaks to the generalizability of the solution.

**Previous Weeks’ Puzzles**

- Can You Solve This Extremely Difficult
*Star Trek*Puzzle? - You Either Solve This Riddle, Or You Die
- Can You Solve ‘The Hardest Logic Puzzle In the World’?
- You’ll Need All 3 Clues To Solve This Puzzle
- Think You Know The Solution To This Classic Riddle? Think Again.
- 100 Lives Are On The Line In This Week’s Puzzle. How Many Can You Save?
- Can You Figure Out This Parking Lot’s Numbering System?
- To Solve This Riddle, Look To Your Family
- Solving This Puzzle Will Help You Grasp The True Nature Of Puzzles
- Can You Guess The Next Number In This Sequence?

## DISCUSSION

The round trip flight without wind will take less time than the round trip with wind. Here is why:

Let’s assume a distance between A and B of 100 miles, an engine speed off 100mph and a wind speed of +/- 10mph.

In the case of no wind, it’s a very simple 200 mile trip, at 100mph, which would take exactly two hours.

In the case of a windy flight, we have to look at both legs separately. If the wind is at your back during the first leg, your absolute speed will be 110mph (1oomph engine speed + 10mph wind speed) and will take 100 miles/110mph = 0.9090909 hours.

The return flight, with the wind in your face will happen at 90mph (1oomph engine speed - 10mph wind speed) and will take 100 miles/90mph = 1.1111111 hours.

If you add these times, you’ll get 2.02020202 hours, which is greater than 2 hours.

A more qualitative proof of this to think what would happen if the wind were moving even faster, say 99.999 mph. This would essentially halve the time to get from A to B (since travel speed has doubled), but will push the return trip’s time to near infinity as the plane’s total speed approaches zero.