A set of dominoes can be arranged into a valid arithmetic sum:

and into a magic square:

From Joseph S. Madachy, *Madachy’s Mathematical Recreations*, 1966, and W.W. Rouse Ball, *Mathematical Recreations and Essays*, 1919.

A set of dominoes can be arranged into a valid arithmetic sum:

and into a magic square:

From Joseph S. Madachy, *Madachy’s Mathematical Recreations*, 1966, and W.W. Rouse Ball, *Mathematical Recreations and Essays*, 1919.

I fire two shots and kill Cristabel. The first bullet strikes her brain, killing her immediately. The second bullet lodges in her heart: it would have killed her, had she not already died because of the first bullet. I argue that I did no serious harm. Bearing in mind what the second bullet would have done, the first bullet merely caused Cristabel the loss of one second of life — hardly serious. The second bullet, of course, did not kill her.

— Peter Cave, *This Sentence Is False*, 2009

5^{3} + 5^{3} = 250

2^{3} + 5^{3} + 0^{3} = 133

1^{3} + 3^{3} + 3^{3} = 55

Australian philosopher J.J.C. Smart asks, “In what units is the rate of time flow to be measured? Seconds per — what?”

- Cain killed a quarter of the world’s population.
- Spencer Tracy’s 1937 Oscar was engraved DICK TRACY.
- 15626 = 1 + 5
^{6×2-6} - NINE TEN ELEVEN alternates vowels and consonants.
- “Can you play chess without the queen?” — Wittgenstein

The cable guy is coming tomorrow between 8 a.m. and 4 p.m. Let’s bet on whether he turns up in the morning or the afternoon.

Both windows are four hours long, so as we sit here today, it seems rational to treat them as equally likely. But suppose you choose the morning. As the clock begins to tick, the morning window will gradually close, making the afternoon seem increasingly preferable. Though your present self regards the two eventualities as equally likely, it seems that your future self won’t. Should that affect your decision today?

(In my experience the guy never turns up at all, so perhaps that solves it.)

Hajek, Alan (2005), “The Cable Guy Paradox,” *Analysis* 65: 112-19.

This card curiosity is attributed to Lewis Carroll.

Lay down eight cards with these values:

Now add the values in each column, find a card of that value in the deck, and place it on top of the lower card. Aces count as 1, jacks as 11, queens as 12, and kings as 13. Thus in the first column 1 + 2 = 3, so you’d place a 3 on top of the 2.

When you’ve done all four columns, repeat the process, placing a 4 on the 3, etc. If a sum is more than 13, subtract 13 from it (for example, Q + 7 = 19 – 13 = 6). When you’ve exhausted the deck you’ll have four kings in the bottom row. Place each of these piles on the card above it, then take up the packs from right to left.

Turn the deck face down again and deal 13 cards in a circle, making note of which was dealt first. Counting from that card, deal 13 more cards, placing them on every second pile (that is, piles 2, 4, 6, etc.) and continuing around the circle until 13 are dealt. Then deal 13 more cards, one onto each third pile (piles 3, 6, 9, etc.), and finish by dealing the last 13 cards, one onto every fourth pile (4, 8, 12, etc.).

Each pile should now contain four cards. Take them up in order, starting with the first.

Now comes the payoff. Spell aloud A-C-E, dealing a card for each letter and turning the last one face up. It will be an ace. Continue with T-W-O, T-H-R-E-E, and so on up through J-A-C-K, Q-U-E-E-N, and K-I-N-G. In each case the last card will have the rank just spelled — and the full count will precisely exhaust the deck.

(From Robert Morrison Abraham, *Winter Nights Entertainments, A Book of Pastimes for Everybody*, 1932.)

Here’s a foolproof way to get anyone to sleep with you. Ask:

- Will you answer this question in the same way that you will answer the next?
- Will you sleep with me?

“If she keeps her word,” writes Richard Mark Sainsbury, “she must answer Yes to the second question, whatever she has answered to the first.”

To multiply 1,639,344,262,295,081,967,213,114,754,098,360,655,737,704,918,032,787 by 71, all you have to do is to place another 1 at the beginning and another 7 at the end.

— Samuel Isaac Jones, *Mathematical Wrinkles*, 1929

Four fourths exceeds three fourths by what fractional part?

“This question will usually divide a company.”

— William Frank White, *A Scrap-Book of Elementary Mathematics*, 1908