You have 12 coins that appear identical. Eleven have the same weight, but one is either heavier or lighter than the others. How can you identify it, and determine whether it’s heavy or light, in just three weighings in a balance scale?

This is a classic puzzle, but in 1992 Washington State University mathematician Calvin T. Long found a solution “that appears little short of magic.” Number the coins 1 to 12 and make three weighings:

First weighing: 1 3 5 7 vs. 2 4 6 8
Second weighing: 1 6 8 11 vs. 2 7 9 10
Third weighing: 2 3 8 12 vs. 5 6 9 11

To solve the problem, note the result of each weighing and assemble a three-digit numeral in base 3 as follows:

Left pan sinks: 2
Right pan sinks: 0
Balance: 1

For example, if coin 7 is light, that produces the number 021 in base 3. Now converting that to base 10 gives 7, the number of the odd coin, and an examination of the weighings shows that it must be light. Another example: If coin 2 is heavy, then we get 002 in base 3, which is 2 in base 10. Note that it’s possible to get an answer that’s higher than 12, e.g. when coin 7 is heavy — in that case subtract the base-10 answer you get from 26.

Another curious method to solve the classic puzzle, this one involving verbal mnemonics, appeared in Eureka in 1950.

(Calvin T. Long, “Magic in Base 3,” Mathematical Gazette 76:477 [November 1992], 371-376.)

09/30/2018 UPDATE: Due to an error in the original paper, the weighings I originally specified don’t work in every case — in the third weighing, the left pan should contain 2 3 8 12, not 1 2 8 12. I’ve amended this in the post above; everything should work now. Sorry for the error; thanks to everyone who wrote in.

A neat observation by Japanese mathematician Jun Maekawa: If an origami model can be flattened without damage, then at any vertex (meeting of edges) in its crease pattern the number of “valley” folds and “mountain” folds always differ by two.

The single-vertex crease pattern above has five mountain folds (folds whose outer surface is colored) and three valley folds (folds whose inner surface is colored). (The fifth mountain fold is a bit hard to notice in this example — it’s folded flat at bottom right.)

One consequence of this is that every vertex has an even number of creases, and therefore that the regions between the creases can be colored with two colors.

Paper folder Toshikazu Kawasaki found a related theorem: At any vertex, the sum of all the odd angles is 180 degrees (and likewise the even).

It was at the place afterwards called Solenhofen. The weather was miserable, as Jurassic weather usually was. The rain beat steadily down, and carbon dioxide was still upon the earth.

The Archaeopteryx was feeling pretty gloomy, for at that morning’s meeting of the Amalgamated Association of Enaliosaurians he had been blackballed. He was looked down upon by the Pterodactyl and the Ichthyosaurus deigned not to notice him. Cast out by the Reptilia, and Aves not being thought of, he became a wanderer upon the face of the earth. ‘Alas!’ sighed the poor Archaeopteryx, ‘this world is no place for me.’ And he laid him down and died; and became imbedded in the rock.

And ages afterward a featherless biped, called man, dug him up, and marvelled at him, crying, ‘Lo, the original Avis and fountain-head of all our feathered flocks!’ And they placed him with great reverence in a case, and his name became a by-word in the land. But the Archaeopteryx knew it not. And the descendant for whom he had suffered and died strutted proudly about the barn-yard, crowing lustily cock-a-doodle-do!

— Samuel P. Carrick Jr., in The Fly Leaf, January 1896

From Lee Sallows:

(Thanks, Lee!)

Given a group of 10 men and 10 women, all straight, is it always possible to pair them off in stable marriages, that is, to pair them so that there exist no man and woman who would prefer each other to the partners they have? Yes:

• In the first round, each man who’s not yet engaged proposes to the woman he most prefers. Then each woman says “maybe” to the suitor she most prefers and rejects all the others. Now she and the suitor she hasn’t rejected are provisionally engaged.
• In each following round, each man who’s not yet engaged proposes to the woman he most prefers and hasn’t yet approached. He does this even if she’s already engaged. Then each woman says “maybe” to her most preferred suitor, even if that means jilting her current provisional fiancé.

This process continues until everyone is engaged (as they must be, since every man must eventually propose to every woman and every woman must accept someone). All the marriages are stable because no man can end up pining for a woman who would prefer him to her own partner — that woman must already have rejected or jilted him at some point during the courting:

In 1962, mathematicians David Gale and Lloyd Shapley showed that stable marriages can always be found for any equal number of men and women.

2⁸⁶ = 77371252455336267181195264

That’s the largest known power of 2 with no zeros in its decimal representation.

Are there higher such powers? No one knows.

I just ran across this, offered by Morton C. Schwartz in an old issue of Pi Mu Episilon Journal:

Take any number of zeros and any number of ones and place them in a circle, in any order. Reproduce the circle a second time, concentrically with the first. Rotate either circle, and any number of places. The number of zeros opposite ones will always be even.

(Morton C. Schwartz, “An Amazing Parity Theorem,” Pi Mu Episilon Journal 5:7 [Fall 1972], 338.)