On a standard calculator keypad like the one shown here, any four-digit number that is typed in a rectangular shape is evenly divisible by 11. Some examples: 7964, 6523, 1793, 7128. (The numbers must not include 0.)

Parallelograms work too: 1562, 6875, 2783, etc.

Discussion and some proofs here.

09/20/2023 More: Take three digits in order from any row, column, or main diagonal and append the same three digits in reverse order (e.g., 951159). The resulting number will always be evenly divisible by 37 (and, indeed, by 1221). Mathematical Gazette, December 1986 and June 1987. See also A Keypad Oddity.

Once I read a book of 100 numbered pages with one sentence on each page. Page 1: ‘The sentence on page 2 is true.’ Page 2: ‘The sentence on page 3 is true.’ And so on to page 100: ‘The sentence on page 1 is false.’

On the second reading, page 100 changes the entire book. If page 1 is false, then the truth is ‘The sentence on page 2 is false.’ Likewise, page 2 becomes ‘The sentence on page 3 is false.’ And so on to page 100, which now should be read as ‘The sentence on page 1 is true.’

What happens on the third reading?

— David Morice, “Kickshaws,” Word Ways 26:1 (February 1993), 44-55. See Yablo’s Paradox.

Can a chess knight visit every square on a board with 4 rows by a series of successive moves?

No, it can’t, and Hungarian mathematics prodigy Louis Pósa proved this while still in his early teens. Suppose that such a tour is possible. Then, on any board bearing the standard checkerboard pattern, the knight will land alternately on white and black squares. But now imagine that the board’s top and bottom rows have been colored red and the middle two rows are blue. Now a knight on any red square must jump to a blue square, and because the board has an equal number of red and blue squares, any knight on a blue square must jump to a red one (if it visits two blue squares in a row, it won’t be able to make up for this later by visiting two red ones in a row). So the knight’s tour on any 4 × n board must alternate strictly between red and blue squares.

“But this is impossible,” notes Colorado College mathematician John J. Watkins. “The same knight’s tour alternated between white and black squares in the one coloring, and between red and blue squares in the other coloring, which would mean the two color patterns must be the same, which of course they aren’t. Isn’t that a clever proof, especially for a teenager to discover?”

(John J. Watkins, Across the Board, 2004. Pósa’s proof is given more rigorously here, and it’s also presented in Ross Honsberger’s 1973 book Mathematical Gems.)

UPDATE: It’s important to note that it’s a “closed” knight’s tour that’s impossible — that’s one that ends where it began. An open tour, which can end anywhere, is possible — it breaks Pósa’s proof because it need not alternate strictly between red and blue squares. Thanks to reader Marjan Milanović for pointing this out.

A late, beautiful contribution by Lee Sallows:

See “Extra Magic” Realized.

(Thanks, Lee!)

This is interesting — in order for a simple arch to stand, its shape when inverted must fit into the limits described by a hanging chain (a catenary).

Spanish Catalan architect Antoni Gaudí followed this principle in designing some of his buildings — he created inverted models and let the shapes of hanging chains or weighted strings determine the shapes of the arches.

In the early 1940s a curious question began to circulate among the members of the Princeton physics department. An ordinary lawn sprinkler like the one shown here would turn clockwise (in the direction of the long arrow) as its jets ejected water (short arrows). If you reversed this — that is, if you submerged the sprinkler in a tank of water and induced the jets to suck in the fluid — would the sprinkler turn in the opposite direction?

The problem is associated with Richard Feynman, who was a grad student at the time (and who destroyed a glass container in the university’s cyclotron laboratory trying to find the answer).

In fact Ernst Mach had first asked the question in an 1883 textbook. The answer, briefly, is no: The submerged sprinkler doesn’t turn counterclockwise because counterbalancing forces at the back of the nozzle result in no net torque. Experiments tend to bear this out, although in some cases the sprinkler turns slightly counterclockwise, perhaps due to the formation of a vortex within the sprinkler body.

I don’t know whether this is contrived or whether a student offered it on an actual exam — Ed Barbeau presented “this little beauty of a howler” in the January 2002 College Mathematics Journal, citing Ross Honsberger of the University of Waterloo in Ontario.