“Editors Have Troubles”

Editors have troubles like less distinguished folk. One of these, who presides over the destinies of a western newspaper, is mourning the loss of two subscribers. One wrote asking how to raise his twins safely, while the other wanted to know how he might rid his orchard of grasshoppers. The answers went forward by mail, but by accident he put them into the wrong envelopes, so that the man with twins received this answer: ‘Cover them carefully with straw and set fire to it, and the little pests, after jumping in the flames a few minutes will be speedily settled.’ And the man with the grasshoppers was told to ‘give them castor oil and rub their gums with a bone.’

— The Typographical Journal, Aug. 15, 1900

(Thanks, Zachary.)

July 31, 2014July 30, 2014 | Oddities

Gear Trouble

A problem from the U.S.S.R. mathematical olympiad:

You’re given 13 gears. Each weighs an integral number of grams. Any 12 of them can be placed on a pan balance, with 6 in each pan, so that the scale is in equilibrium. Prove that all the gears must be of equal weight.

	Select Click for Answer
Each gear weighs either an even or an odd number of grams. Any set of 12 gears can be divided into two groups of equal weight, so a set of 12 gears has a total weight that’s even. This weight remains even if one of the 12 gears is exchanged with the 13th gear, and this is true no matter which of the 12 is exchanged. So either all the gears are of even weight or all are of odd. Now find the weight of the lightest gear and subtract it from all 13 gears. This gives us a set of 13 gears that still fulfill the conditions of the problem (though one or more of them now has zero weight). But now we know that each gear has an even weight: If all had originally been even, then we’ve subtracted an even number to produce an even number. And if all were originally odd, then we’ve subtracted odd from odd to produce even. Now, since each gear has even weight, we can divide the weight of each gear by 2 to obtain another set of 13 gears that still fulfills the conditions of the problem. If all the original gears had the same weight, we can continue to divide by 2 repeatedly without a problem: With each division, we produce a new set that satisfies the conditions, and every gear’s weight will have the same parity, even or odd. But if the original group of gears contained gears of differing weights, then repeated division by 2 will eventually produce a situation in which some weights are even and some odd. No such set can satisfy the original conditions of the problem. So all the original gears must have had the same weight.

Click for Answer

July 31, 2014August 10, 2014 | Puzzles

Niche Publishing

The Epworth Instigator, a monthly publication in Santa Monica, edited by Saml. Carlisle, has probably the smallest sworn circulation statement of any paper in the United States. According to the sworn statement, Forrest Harris, the business manager, says that the number of copies printed and circulated for the month of August, 1907, was one.

The paper is published in the interests of the Epworth league here, and the only copy is taken to the meeting and read aloud, advertisements and all.

— Printers’ Ink, Oct. 16, 1907

(Thanks, Craig.)

July 30, 2014July 29, 2014 | Literature

Parting of the Waters

In Wyoming’s Teton Wilderness, North Two Ocean Creek splits into two smaller creeks: Pacific Creek flows westward to the Pacific Ocean, Atlantic Creek eastward to the Atlantic.

In principle this makes it possible for a fish to swim upstream from the Pacific, cross the Great Divide, and pass down into the Atlantic. This may be how the Yellowstone cutthroat trout came to be found in both the Snake and Yellowstone rivers.

July 30, 2014July 30, 2014 | Trivia

Looking Up

Rome is north of Chicago.

Venice is north of Minneapolis.

London is north of Calgary.

Paris is nine miles south of the U.S.-Canada border.

July 29, 2014July 28, 2014 | Trivia

Muse Be Damned

Anthony Trollope established himself as one of the world’s most prolific novelists while holding down a 30-year career as a full-time civil servant.

He did this by simply demanding it of himself. Even while traveling he rose at 5:30 each morning and worked for three hours, “allowing himself no mercy,” counting words as he went and noting his progress on a chart, “so that if at any time I have slipped into idleness for a day or two, the record of that idleness has been there, staring me in the face, and demanding of me increased labour, so that the efficiency might be supplied.” He disdained inspiration: “To me it would not be more absurd if the shoemaker were to wait for inspiration, or the tallow-chandler for the divine moment of melting.”

“All those I think who have lived as literary men — working daily as literary labourers — will agree with me that three hours a day will produce as much as a man ought to write,” he wrote in his autobiography. “But then he should so have trained himself that he shall be able to work continuously during those three hours — so have tutored his mind that it shall not be necessary for him to sit nibbling his pen, and gazing at the wall before him, till he shall have found the words with which he wants to express his ideas. It had at this time become my custom … to write with my watch before me, and to require from myself 250 words every quarter of an hour. I have found that the 250 words have been forthcoming as regularly as my watch went.”

His brother Tom said, “Work to him was a necessity and a satisfaction. He used often to say he envied me the capacity for being idle.”

July 29, 2014July 28, 2014 | Literature

Podcast Episode 20: Life Imitates Science Fiction

In 1944, fully a year before the first successful nuclear test, Astounding Science Fiction magazine published a remarkably detailed description of an atomic bomb. The story, by the otherwise undistinguished author Cleve Cartmill, sent military intelligence racing to discover the source of his information — and his motives for publishing it.

In this episode of the Futility Closet podcast we’ll follow the investigation that ensued, which involved legendary editor John W. Campbell and illuminated the imaginative power of science fiction and the role of censorship in times of war. We’ll also hear Mark Twain’s advice against being too clever and puzzle over the failure of a seemingly perfect art theft.

See full show notes …

July 28, 2014November 7, 2017 | Podcast

In a Word

obolary
adj. poor or impecunious

inopious
adj. lacking wealth or resources; needy

ptochocracy
n. government by the poor

July 28, 2014July 27, 2014 | Language

Shadow Play

Artists Tim Noble and Sue Webster find self-portraits in arrangements of domestic trash. Their Dirty White Trash (with Gulls) (1998, right) was contrived from the six months’ rubbish they produced while making it, a sculpture produced by the residue of its own composition.

More shadow art: Shigeo Fukuda, Larry Kagan, Richard Haas.

lead pencil billboard

Somewhat related: Last year travelers from Washington state to Vancouver were surprised to discover this “negative space” billboard by the side of the road. It was created by Daniel Mihalyo and Annie Han of Seattle-based art collective Lead Pencil Studio. Most billboards draw the eye away from the environment; this draws the eye to it.

(Thanks, Alex and Bob.)

July 27, 2014July 26, 2014 | Art

Coming and Going

A puzzle by Pierre Berloquin:

In my house are a number of rooms. (A hall separated from the rest of the house by one or more doors counts as a room.) Each room has an even number of doors, including doors that lead outside. Is the total number of outside doors even or odd?

	Select Click for Answer
Call a door internal if it opens between rooms or external if it opens outside. And designate each room either O (if it has an odd number of internal doors) or E (if even). Each E room has an even number of external doors, since all rooms have an even number of doors and even – even = even. So the E rooms, taken as a group, have a total number of external doors that’s even. The total number of O rooms is either odd or even. If it’s odd, then the total number of doors in the O rooms is odd, since the sum of an odd number of odd numbers is odd. But this is impossible: Every internal door must open on two rooms, and this would leave one door that has no second room to open on. So the total number of O rooms is even. Since each O room has an odd number of internal doors and an even number of total doors, each must have an odd number of external doors. Because the total number of O rooms is even, this means that the O rooms taken as a group have an even number of external doors (since the sum of an even number of odd numbers is even). Since both groups, the E rooms and the O rooms, have an even number of external doors, so does the whole house. L.A. Graham provided a simpler solution in 1959.

Click for Answer

July 26, 2014August 3, 2014 | Puzzles