A great portrait is always more a portrait of the painter than of the painted. When we look at a portrait by Holbein or Rembrandt it is of Holbein or Rembrandt that we think more than of the subject of their picture. Even a portrait of Shakespeare by Holbein or Rembrandt could tell us very little about Shakespeare. It would, however, tell us a great deal about Holbein or Rembrandt.
A little kingdom contains 66 people, a king and 65 citizens. Each of them, including the king, has a salary of one gold piece. When democracy comes, the king is denied a vote, but he has the power to suggest changes, in particular regarding the redistribution of salaries. The salaries must total 66, and each salary must be a whole number of gold pieces. The citizens will vote on each suggestion, which will pass if more citizens vote for it than against it. Each voter will reliably support a measure if it will increase his salary, oppose it if it will decrease his salary, and otherwise abstain from voting.
The king is greedy. What’s the highest salary he can arrange for himself?
Impressively, he can get 63 gold pieces. He starts by proposing that just over half of the 65 citizens (33 voters) have their salaries doubled to 2 gold pieces, at the expense of the other 33 voters, including himself. Then he does the same thing again, proposing that just over half (17) of the 33 salaried voters receive an increase to 3 or 4 gold pieces, while the remaining 16 in that group are reduced to zero. By continuing in this manner he can reduce the number of salaried voters to 9, 5, 3, and finally 2, with each receiving 33 gold pieces. Then the king can propose that three of the unsalaried citizens receive a salary of 1 gold piece if these two large salaries are now reassigned to himself.
From Peter Winkler’s excellent Mathematical Puzzles, 2021.
Now this last example is very interesting. What happened when Christ broke the bread at the Last Supper? He broke a thing into fragments. But each piece contained the whole: that is, his entire body. And, again, what happens if the Host, after it has been consecrated by a priest, falls to the floor of the church and crumbles, and if a mouse then eats the crumbs? Does the mouse eat Christ’s body? I do not wish to develop this argument, merely to remind you that this was one of the arguments used by Protestant reformers to ridicule Catholic practice.
— Jacqueline Lichtenstein, “The Fragment: Elements of a Definition,” in William Tronzo, ed., The Fragment: An Incomplete History, 2009
One night in 1939, Wolcott Gibbs’ 4-year-old son Tony began chanting a song in the bathtub. It was sung “entirely on one note except that the voice drops on the last word in every line”:
He will just do nothing at all.
He will just sit there in the noonday sun.
And when they speak to him, he will not answer them,
Because he does not care to.
He will stick them with spears and throw them in the garbage.
When they tell him to eat his dinner, he will just laugh at them.
And he will not take his nap, because he does not care to.
He will not talk to them, he will not say nothing.
He will just sit there in the noonday sun.
He will go away and play with the Panda.
He will not speak to nobody because he doesn’t have to.
And when they come to look for him they will not find him.
Because he will not be there.
He will put spikes in their eyes and put them in the garbage.
And put the cover on.
He will not go out in the fresh air or eat his vegetables.
Or make wee-wee for them, and he will get thin as a marble.
He will do nothing at all.
He will just sit there in the noonday sun.
Pete Seeger liked this so much that he made a song of it — he called it “Declaration of Independence”:
Here’s a unit square. Prove that, if nine points are identified in the square’s interior, we can always find three of them that form a triangle of area 1/8 or less.
Cut the square into four smaller squares, each of which has a side length of 1/2. Now one of these smaller squares must contain at least three of the points. And the largest triangle that these points can form has an area half that of the small square — or 1/8.
05/16/2024 This is an old Martin Gardner puzzle, intended to illustrate the pigeonhole principle. Reader Jon Jerome points out that the three points in the small square might be collinear, but if we accept a degenerate triangle with zero area, then the proof holds.
06/02/2024 Reader Drake Thomas writes:
“An alternate approach to nine points in the square: sort the points by y coordinate. There are four disjoint intervals [y1, y3], [y3, y5], [y5, y7], [y7, y9], so at least one of those intervals [yi, yi+2] is of length at most 1/4. Then the triangle yi, yi+1, yi+2 occupies at most half of the rectangle of dimensions 1 × 1/4, so is area at most 1/8.”
Hong Kong contains a street named Rednaxela Terrace. It’s hard not to notice that this is Alexander spelled backward, but the origin of the name is uncertain.
In Signs of a Colonial Era (2009), Andrew Yanne and Gillis Heller claim that the street had been named Alexander Terrace after its original owner but that a clerk recorded the name backward, as the Chinese language was written right to left at the time.
Another possibility is that the name is linked to New York abolitionist Robert Alexander Young’s 1829 pamphlet Ethiopian Manifesto, which contains the name Rednaxela.
In writing novels as well as plays the cardinal rule is to treat the various characters as if they were chessmen, and not try to win the game by altering the rules; for instance, not move the knight as if it were a pawn, and so on. Again the characters ought to be strictly defined, and not put out of action in order to help the author to accomplish his purpose; for, on the contrary, it is through their activity alone he should try to win. Not to do this is to appeal to the miraculous, which is always unnatural.
Two numbers are mirror numbers if each presents the digits of the other in reverse order, such as 123 and 321. Find two mirror numbers whose product is 92,565.
This solution is by V. Dubrovsky. The size of the product shows that the factors must have three digits each. So let one of them be abc (or 100a + 10b + c) and the other be cba. The product ends in 5, so either a or c must be 5. Say that’s a. The other factor starts with 5, and 92,565 / 500 < 200, so c must be 1. As to b, we can see that the 6 in 92,565 is the last digit of 5b + b, or 6b, so b must be either 1 or 6, and we can test these candidates to learn that it’s 6. The numbers we seek are 165 and 561.
05/17/2024 UPDATE: Reader Robert Filman points out that we can solve this without actually having to multiply the numbers together. The sum of the digits of any number is the remainder of that number divided by 9, and the sum of the digits of 92,565 mod 9 = 0. So once we’ve established that the end digits of the factors we’re seeking are 1 and 5 and that the middle digit is 1 or 6, as above, we can notice that none of the resulting candidates (115, 165, 511, 561) has a digit sum divisible by 9 and hence each factor must be divisible by 3 — which means that the digit sum of each factor must be a multiple of 3. The only possibilities are 165 and 561. (Thanks, Robert.)