In the classic Indian rope trick, a rope rises into the sky, its end lost to view. A boy disappears up the rope, and when he fails to return the angry magician climbs up after him. Body parts fall to the ground, the magician descends and places the parts in a basket, and the boy reappears uninjured.

This is all thought to be a legend, but in 1979 mathematician J.L.G. Pinhey of The Perse Boys’ School worked out that levitating a rope is possible, at least in principle. If the top of the fakir’s rope is 1.5 × 10⁸ meters above Earth’s surface, it will simply stand erect, its position sustained by the motion of the planet.

“Since the rope between its ends is in tension the configuration is stable, and the faqir and his boy-victim can climb it in safety. However, in order to drop the bits to earth, the pair must not climb even a quarter of the way to the top.”

(J.L.G. Pinhey, “63.12 The Indian Rope Trick,” Mathematical Gazette 63:424 [June 1979], 110-111.)

From Wikimedia user Cmglee, two digital arithmetic techniques:

Above: To multiply a positive single-digit integer by 9, hold up your hands palm up, imagine the fingers numbered consecutively 1 to 10, and fold down the finger corresponding to the number to be multiplied (here, 8). The product is the two-digit number represented by the remaining two groups of fingers — here there are seven fingers to the left of the folded finger and 2 to the right, so 9 × 8 = 72.

Below: To multiply two integers between 6 and 10, imagine each hand’s fingers numbered from 6 (pinky) to 10 (thumb), as shown. Fold down the two fingers corresponding to the factors, as well as all fingers between these two (in this example we’ll calculate 6 × 7, so fold down finger 7 on the left hand, finger 6 on the right, and the finger that lies between them, the left pinky). Count the remaining upraised fingers on the left hand (3), multiply that by the remaining upraised fingers on the right hand (4), and add 10 times the number of folded fingers (30). 3 × 4 + 30 = 42.

A square can be divided into an even number of triangles of equal area — but not into an odd number.

Suppose your dog eats the ace of spades and you’re forced to play poker with a 51-card deck. Does this make it more or less likely that you’ll be dealt a two-pair hand?

The answer is not immediately clear. The loss of the ace means that fewer two-pair hands are possible, but the total universe of possible five-card hands is also reduced.

Surprisingly, the probability is unchanged. “[T]he probability of getting a Two-Pair hand in the 51-card deck with the missing Ace of Spades is exactly the same as for a 52-card standard deck,” notes George Mason University mathematician D.M. Anderson. And “[i]t turns out that this seemingly magical result for a deck with a missing Ace of Spades is not limited to Two-Pair hands. Indeed, examining the traditional hierarchy of 5-card poker hands, we find an analogous result holds for One-Pair hands, Three-of-a-Kind, Full Houses and Four-of-a-Kind.”

(D.M. Anderson et al., “Not Playing With a Full Deck?”, Recreational Mathematics Magazine 11:18 [March 2024], 51-63.)

This remarkable phenomenon was discovered by Cambridge mathematician James Grime. Number five six-sided dice as follows:

A: 2, 2, 2, 7, 7, 7
B: 1, 1, 6, 6, 6, 6
C: 0, 5, 5, 5, 5, 5
D: 4, 4, 4, 4, 4, 9
E: 3, 3, 3, 3, 8, 8

Now, on average:

A beats B beats C beats D beats E beats A

and

A beats C beats E beats B beats D beats A.

Interestingly, though, if each die is rolled twice rather than once, then the first of the two chains above remains unchanged except that D now beats C — and the second chain is reversed:

A beats D beats B beats E beats C beats A.

As a result, if each of two opponents chooses one of the five dice, a third opponent can always find a remaining die that beats them both (so long as he’s allowed to choose whether the dice will be rolled once or twice).

(Ward Heilman and Nicholas Pasciuto, “What Nontransitive Dice Exist Among Us?,” Math Horizons 24:4 [April 2017], 14-17.)