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.)

This is easiest if we attack it from the end. When there are just two pirates left, P2 and P1, P2 should obviously propose that he keep all the gold for himself and give none to P1. The two of them will vote, and P2 will carry 50 percent of the vote and come away 100 gold pieces richer.

Now back up a step and consider the case with three pirates. P1 knows that if he doesn’t accept P3’s proposal then the situation will revert to the one we just considered, where he comes away empty-handed. So he should accept any proposal that gives him more than nothing. P3, who’s smart enough to see this, should propose giving 1 gold piece to P1 and nothing to P2.

Back up and consider four pirates. P4 will vote for his own proposal, of course, but he needs the vote of one other pirate to reach 50 percent, and he wants to spend as little gold as possible to get it. He proposes 99 gold pieces for himself, 0 for P3, 1 for P2, and 0 for P1. P2 is glad to vote for this, because if he doesn’t, P4 will go to the sharks and we’ll revert to the case of three pirates, where P2 gets nothing.

Back up one more step to five pirates. P5 needs two other pirates to vote for his proposal in order to have 50 percent of the votes. He proposes 98 gold pieces for himself, 0 for P4, 1 for P3, 0 for P2, and 1 for P1. P3 and P1 will vote for this, because otherwise they’ll find themselves in the four-pirate scenario, where they get nothing.

You can see where this is headed. In the original scenario, with ten pirates, P10 should propose 96 gold pieces for himself; 1 for each of the pirates P8, P6, P4, and P2; and none for the rest. By allocating just 4 gold pieces cannily, he can keep nearly all the treasure.

(Ian Stewart, “A Puzzle for Pirates,” Scientific American, May 1999, 98-99.)

Reduce the river to a line of zero width along the bank nearest to A, and imagine moving City B up correspondingly to B’. Now a straight line drawn between A and B’ shows us where to put the bridge.

To prove this we can appeal to the triangle inequality, which says that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side. In the diagram below, the optimum path we’ve suggested goes A-P-Q-B. A bridge will have the same length no matter where we build it, so let’s omit that segment from our thinking. QB is parallel to PB’, so the length of our proposed road A-P-Q-B, minus the width of the river, is equivalent to AB’.

Now consider any alternative road, say A-P’-Q’-B. Again we can ignore the width of the river, and Q’B is parallel to P’B’, so the length of the alternative road A-P’-Q’-B, minus the width of the river, is equivalent to A-P’-B’. The triangle inequality says that AB’ must be shorter than AP’ + P’B’, so A-P-Q-B must be shorter than A-P’-Q’-B. We can apply a similar argument for any other proposed site for the bridge — bridge PQ must always produce a shorter road.

From Zbigniew Michalewicz and David B. Fogel, How to Solve It: Modern Heuristics, 2000.

The black rook on d4 pins the white rook on d6, so the black bishop on b6 is free to move, and this means that the white knight on c7 is pinned and not checking the black king. White plays 1. f3, threatening to take the e2 pawn with his queen. This would be mate — it’s no use blocking the queen’s check with the d4 rook, because this reverses the chain of pins — now the pinned d3 rook doesn’t check, so the d6 rook does, the b6 bishop doesn’t, and the c7 knight does, giving mate anyway! Likewise:

1. f3 Rd5 2. Qxe2# (same idea as above — 2. … Rb5 is ineffective)
1. f3 Bxa5 2. Kc8#
1. f3 Bxc7 2. Nxc7#
1. f3 Bxe8 2. Kxe8#
1. f3 Qxe7+ 2. Kxe7#
1. f3 Rd2 2. Bxd2#
1. f3 Rxd6+ 2. Qxd6#

Dawson wrote, “The idea amounts to ‘discovering mate’ from c7.” He dedicated the puzzle to Eduard Birgfeld. From Caissa’s Wild Roses, 1935.