The game must have reached this position:

Now:

1. … Bd5+ 2. c4 bxc3+ e.p. 3. Kxc3+

This is the only legal way that the given position could have been reached. Therefore the king must have been knocked from c3.

“Since the identical counterfeit bill can be traced through all the transactions, these are all invalid. Therefore everybody stands in relation to his debtor just where he was before the banker picked up the note, except that the butcher owes, in addition, $5.00 to the farmer for the calf received.”

Update: A number of perceptive people have written in questioning Dudeney’s solution, and after some thought I’m inclined to agree with them. The intermediate transactions can stand as they are: Each participant has lost $5 (by accepting a fake bill) and gained $5 (by using it to pay a debt). But the banker has gained $5 (by finding and spending a bad bill), and the bank has lost $5 (by accepting it as good). So if the banker now pays $5 to the bank, all is well.

This puzzle first appeared in The Strand in 1917, and it’s appeared in various collections since then, but I have not seen these objections raised until now. This means I have the best readers in the world.

If no two people are to reach the same handshake total, then the most gregarious guy in the stadium must shake 49,999 hands, the next-most-gregarious 49,998, and so on. But that means the shyest guy would shake zero hands — which is impossible, since the gregarious guy shook every hand in the place.

Alternatively, if the shyest guy shakes one hand, then to avoid any duplication the gregarious guy would have to shake 50,000 hands, which is also impossible, as there are only 49,999 hands available for him to shake. Therefore there must be at least two participants, somewhere, who have shaken the same number of hands.

This is an example of the “pigeonhole principle” — which also explains why 30,000 of us have the same number of hairs on our heads.

1. Qe2. If the e3 pawn makes a capture, then Qxe4 is mate. If the black king takes a pawn, then the queen gives mate on the fifth rank.

From D. Littlewick, “Curiosities on the Chess-Board,” Strand, January 1910.

1. Kc5 Kd7 2. Kb6 Ke7 3. Kb7 Kd7 4. Kb8 Ke7 5. Kc8 Kf8 6. Kd7#:

From the Huddersfield College Magazine, 1877.

04/10/2014 Wait, this doesn’t work — Black has 4. … Kd8. A better solution is 1. Kc5 Kd7 2. Kb6 Ke7 3. Ka7 Kd7 4. Kb7 Ke7 5. Kc8 Kf8 6. Kd7# (Thanks, Stanley.)

The next number is 9.

The numbers correspond to the Roman numerals hidden in English number names:

Solution to Road Work:

I run the first half-mile in 4 minutes and the next in 5 minutes, and continue alternating like this throughout the marathon. In this way I’ll cover any measured mile in 9 minutes. But because the “fast” segments outnumber the “slow” ones, my average pace for the whole run is less than 9 minutes per mile. This works because the course does not cover an even number of miles.

From Frank Morgan, The Math Chat Book, 2000.

There are two possible answers here, thanks to the slipperiness of the word all:

1. I have a dog, a cat, and a fish.

2. I have two Denebian Slime Devils, or two Eight-Thumbed Mongolian Rat-Weasels, or two of anything-that’s-not-a-cat-dog-or-fish.

I actually prefer the second answer. (Thanks, Remy.)

(a) a8, (b) h1, (c) e3, (d) g7.