The men sold x cows for $x, so their income from that sale is $x². x² isn’t divisible by 12, so x can’t be a multiple of 6 (if it were, then x² would be divisible by 36 and hence by 12). So x is 12n ± y, where n is any integer and y is an integer from 1 to 5.

Each man ends up with the same number of animals, so the total number of animals must be even, which means there’s an odd number of sheep and one lamb. So (12n ± y)² when divided by 12 should leave a remainder and have an odd integer for its quotient.

12n² ± 2ny will always be even, so that last term, y²/12, must provide an odd contribution toward this quotient. In order to do this, y² must exceed 12, so y must be 4 or 5. If it’s 5, we get

which provides an even contribution (2) to the quotient. So y must be 4 and y² = 16:

That’s what we need; the integer part of this quotient, 1, is odd. Since the remainder when 4² is divided by 12 is 4, a lamb costs $4. Consequently, in dividing the flock one man got a $4 lamb and the other got a $12 sheep. To make this fair, the traded harmonica must be worth $4.

(The problem was proposed by J.H. Neelley, T.L. Smith, and P.S. Ganesa Sastri in AMM 37:3, 162; Ross Honsberger introduced the harmonica in recounting it in his Mathematical Morsels, 1978. This solution is by P.S. Ganesa Sastri of Trichinopoly, South India.)

All that is necessary is to add the two distances at which they meet to twice their difference. Thus 720 + 400 + 640 = 1760 yards, or one mile, which is the distance required.

A more general solution can be found on page 416 of Pi Mu Epsilon Journal, Spring 1982 (PDF).

Set all the variables equal to 2! “The point of the distributive law is that its application doesn’t change the value of the expression,” writes Dartmouth mathematician Peter Winkler. “The value of the initial expression limits the size of anything you can get from it by expansion.”

Winkler received the puzzle from Dick Lipton of the College of Computing at Georgia Tech and included it in his “Seven Puzzles You Think You Must Not Have Heard Correctly” at the seventh Gathering for Gardner, in March 2006.

The first rook goes on d2. Now if the king flees to the left, check him from the a-file, and if he flees to the right, check him from the g-file. So, for example, if he goes to e4, then play 2. g4+:

That’s the key — this plan traps the king in a narrow pair of files, and now it’s an easy matter to attack those with the remaining rooks (above, for example: 2. … Kf3 3. Rf4+ 4. Ke3 Re2#).

From Chess Braintwisters, 1999.

In long division, whenever two digits are brought down instead of one — as happens twice in this problem — a zero must appear in the quotient. That tells us that the quotient is x080x.

The visible 8 multiplied by the divisor produces a three-digit number. But when the last number in the quotient is multiplied by the divisor, it produces a four-digit number. So the last digit in the quotient must be 9.

The divisor has three digits and produces a three-digit number when multiplied by the visible 8. So the divisor must be less than 125, because 8 × 125 = 1000.

That tells us something further. When the divisor is multiplied by the first digit of the quotient, it produces a three-digit number, and when this is subtracted from the first four digits of the dividend it yields a two-digit difference. That means that the first digit of the quotient must be greater than 7, because 7 × 124 = 868, and that would yield a three-digit difference when subtracted from even the smallest possible four-digit number, 1000 (1000 – 868 = 132). An 8 or a 9 could produce a two-digit difference, but a 9 would produce a four-digit number when multiplied by the divisor. So the first digit of the quotient is 8, and the full quotient is 80809.

Now, 80809 × 123 = 9939507, which has only seven digits, and the dividend has eight. So the divisor is less than 125 and greater than 123, which means it’s 124. That gives us everything we need:

“The portrait may be drawn in a single line because it contains only two points at which an odd number of lines meet, but it is absolutely necessary to begin at one of these points and end at the other. One point is near the outer extremity of the King’s left eye; the other is below it on the left cheek.”

From the road, Marsh alone walked to the edge of the cliff, carrying Lamson’s boots in his hands. At the edge of the cliff he changed boots and walked backward to the road, carrying his own boots.

“This little maneuver accounts for the smaller footprints showing a deeper impression at the heel, and the larger prints a deeper impression at the toe, for a man will walk more heavily on his heels when going forward, but will make a deeper impression with the toes in walking backwards. It will also account for the fact that the large footprints were sometimes impressed over the smaller ones, but never the reverse; also for the circumstance that the larger footprints showed a shorter stride, for a man will necessarily take a smaller stride when walking backwards.”

From Henry Dudeney’s The Canterbury Puzzles, 1907.

Review reader Brad Edelman offered this solution. First number the tiles 0 to 6, left to right, with pieces in the same column numbered in sequence from top to bottom. Now:

2 – down, down
4 – down
6 – right
3 – down, right, down, down
1 – right, right
2 – up, up, left
4 – left, down
3 – left
1 – down, right, down
2 – right, right
4 – up, up
3 – left
2 – down
5 – left
6 – up
1 – right
2 – right
3 – right
4 – right
0 – right, right, down, down
5 – left, left, left, down, down
4 – up, left, left, left, up

The original puzzle was posed in the November/December 2007 issue; the solution was presented two issues later.

At the time, readers made some interactive tools, in Word and Flash, to aid in the solving, but those don’t seem to be available any longer. If anyone knows of any, please let me know and I’ll add a link here.

12/06/2016 UPDATE: Reader Chuck Rubins has created a downloadable interactive version in PowerPoint. (Thanks, Chuck.)