Strategies to memorize π sometimes rely on devising a sentence with words of representative lengths. Isaac Asimov offered this one:

How I want a drink, alcoholic, of course, after the heavy lectures involving quantum mechanics!

Count the letters in each word and you get 3.14159265358979. That’s not the best strategy, though: What shall we do about zero? And it’s hard to reel off the digits impressively for friends when you have to stop and count the letters in each word.

Arthur Benjamin, a mathematician at Harvey Mudd College, suggests using a phonetic code instead:

1 = t or th or d
2 = n
3 = m
4 = r
5 = l
6 = sh, ch, or j
7 = k or hard g
8 = f or v
9 = p or b
0 = s or z

Once we’ve memorized this list, we can turn a whole string of digits into a single word simply by inserting vowels between the consonants. So, for example, 314 could be represented by meter, motor, meteor, matter, mother, etc. The consonants h, w, and y, which don’t appear in the list, can be used like additional vowels.

Using this method, it takes only three sentences to encode the first 60 digits of π:

3. 1415 926 5 3 58 97 9 3 2 384 6264
My turtle Pancho will, my love, pick up my new mover Ginger.

3 38 327 950 2 8841 971
My movie monkey plays in a favorite bucket.

69 3 99 375 1 05820 97494
Ship my puppy Michael to Sullivan’s backrubber!

Now we just need a way to remember the sentences …

(Arthur T. Benjamin, “A Better Way to Memorize Pi: The Phonetic Code,” Math Horizons 7:3 [February 2000], 17.)

The invention of logarithms came on the world as a bolt from the blue. No previous work had led up to it, foreshadowed it, or heralded its arrival. It stands isolated, breaking in upon human thought abruptly without borrowing from the work of other intellects or following known lines of mathematical thought.

— Lord Moulton on the 300th anniversary of John Napier’s 1614 book Description of the Wonderful Canon of Logarithms. E.W. Hobson called the invention “one of the very greatest scientific discoveries that the world has seen.”

Usain Bolt is such a great sprinter that his distinctions may extend to other worlds.

In 2013, University of Leicester physics undergraduate Hannah Lerman and her colleagues determined that the Jamaican athlete was one of the few humans who could get aloft on Saturn’s moon Titan with wings strapped to his arms.

Factoring in Titan’s gravity and atmospheric density, Lerman found that a person could take flight in a normal-sized wingsuit only if they could run at 11 meters per second.

“This speed has been reached but only by the fastest human runners, for example, Usain Bolt, who ran almost 12 m/s,” Lerman wrote. “For an average human to take off with the standard wingsuit they would require some sort of propulsion device to give them enough speed to take off.”

(H. Lerman, B. Irwin, and P. Hicks, “P5_1 You Can Fly,” Journal of Physics Special Topics, University of Leicester, Oct. 22, 2013.)

A paradox by the German mathematician Martin Löb:

Let A be any sentence. Let B be the sentence: ‘If this sentence is true, then A.’ Then a contradiction arises.

Here’s the contradiction. B makes the assertion “If B is true, then A.” Now consider this argument. Assume B is true. Then, by B, since B is true, A is true. This argument shows that, if B is true, then A. But that’s exactly what B had asserted! So B is true. And therefore, by B, since B is true, A is true. And thus every sentence is true, which is impossible.

(Lan Wen, “Semantic Paradoxes as Equations,” Mathematical Intelligencer 23:1 [December 2001], 43-48.)

In the January 2001 issue of the College Mathematics Journal, University of Calgary mathematician Jonathan Schaer offers this simple proof that arctan 1 + arctan 2 + arctan 3 = π. The sum of the angles of this large triangle is 180°. And the diagram shows that its lower left angle is arctan 3, its lower right angle is arctan 2, and its top angle is part of an isosceles right triangle. So arctan 1 + arctan 2 + arctan 3 = 180°, or π radians. “No words, even no symbols!”

(“Miscellanea,” College Mathematics Journal 32:1, 68-71.)

Welsh mathematician Robert Recorde’s 1543 textbook Arithmetic: or, The Ground of Arts contains a nifty algorithm for multiplying two digits, a and b, each of which is in the range 5 to 9. First find (10 – a) × (10 – b), and then add to it 10 times the last digit of a + b. For example, 6 × 8 is (4 × 2) + (10 × 4) = 48.

This works because (10 – a)(10 – b) + 10(a + b) = 100 + ab, and it saves the student from having to learn the scary outer reaches of the multiplication table — they only have to know how to multiply digits up to 5.

(From Stanford’s Vaughan Pratt, in Ed Barbeau’s column “Fallacies, Flaws, and Flimflam,” College Mathematics Journal 38:1 [January 2007], 43-46.)

Iowa State University mathematician Alexander Abian was a quiet man with a bold idea: He believed that blowing up the moon would solve most of humanity’s problems. In thousands of posts on Usenet, he maintained that destroying the moon would eliminate Earth’s wobble, canceling the seasons and associated calamities such as hurricanes and snowstorms.

“You make a big hole by deep drilling, and you put there atomic explosive,” he explained in 1991. “And you detonate it — by remote control from Earth.”

“I was questioned about it,” wrote English astronomer Patrick Moore in Fireside Astronomy (1993). “I pointed out, gently, that even if the Moon were removed it would not alter the tilt of the Earth’s axis in the way that the professor seems to believe. Moreover, the energy needed to destroy the Moon would certainly destroy the Earth as well, even if we had the faintest idea of how to do it. The British Meteorological Office commented that a moonless Earth would be ‘bleak and tideless’, and a spokesman for the British Association for the Advancement of Science, struggling nobly to keep a straight face, asked what would happen if the experiment went wrong. Predictably, Professor Abian was unrepentant. ‘People don’t seem prepared to sacrifice the Moon for a better climate. It is inevitable that the genius of man will one day accept my ideas.'”

For better or worse, he maintained this position until his death in 1999. “I am raising the petulant finger of defiance to the solar organization for the first time in 5 billion years,” he said. “Those critics who say ‘Dismiss Abian’s ideas’ are very close to those who dismissed Galileo.”

In 2011, Monash University mathematician Burkard Polster set out to answer a practical question: How precariously can you place a laptop computer on a crowded bedside table so that it will take up minimal space without falling off?

Assuming that both the table and the laptop are rectangular, and that the laptop’s center of gravity is its midpoint, it turns out that the optimal placement occurs when the laptop’s midpoint coincides with one of the table’s corners and the footprint is an isosceles right triangle, as above.

This also assumes that the table is reasonably sized. But then, if it’s tiny, then balancing a laptop on it probably isn’t your biggest problem.

(Burkard Polster, “Mathematical Laptops and Bedside Tables,” Mathematical Intelligencer 33:2 [July 2011], 33-35.)

A book lover is thinking of buying six books from a group of eight. The price of each book is a whole number of dollars, and none is less than $2. The prices are such that each possible selection of six books would cost the buyer a different sum. In the end he can’t make up his mind and buys all eight books. What is the smallest amount he must pay?

Each choice of six books from the group of eight leaves two behind. The price of each possible omitted pair must be unique, or else the corresponding sextets would cost the same, which we know is not the case. So we can solve the problem by working out the lowest possible price for each book that ensures that every possible pair has a distinct price.

The three lowest possible prices are $2, $3, and $4. That’s fine so far, but the next book can’t cost $5, because then we’d have two pairs with the same value (5 + 2 = 3 + 4). So we jump to 6 and see if that works. By looking always for the smallest possible next higher price for each volume, we’ll arrive at 2, 3, 4, 6, 9, 14, 22, 31, which gives a total price of $91.

But, interestingly, that’s not the answer! After Roland Sprague published this puzzle in his 1963 book Recreation in Mathematics, he found the solution 2, 3, 4, 6, 10, 15, 20, 30, which totals 90. And Fritz Düball later found 2, 3, 4, 6, 10, 16, 21, 26, which totals 88. Is that the lowest sum possible? Sprague doesn’t claim that it is, and I’ve not seen this problem elsewhere than in his book.

06/19/2022 UPDATE: Reader Michael Küll wrote a program to search for all solutions in a reasonable range. There are four solutions with a total amount less than or equal to $91:

88 :   2   3   4   6   10   16   21   26
90 :   2   3   4   6   10   15   20   30
91 :   2   3   4   6    9   14   22   31  
91 :   2   3   4   6   10   17   22   27

So Düball’s solution is indeed the best possible. (Thanks, Michael.)