In David Hayes and Tatiana Shubin’s Mathematical Adventures (2004), University of California-Davis mathematician Don Chakerian describes a method used in antiquity for solving an equation in one unknown. He illustrates it with a problem from Daboll’s Schoolmaster’s Assistant (1800):

A, B, and C built a house which cost $500, of which A paid a certain sum, B paid 10 dollars more than A, and C paid as much as A and B both; how much did each man pay?

We’ll make two guesses as to how much A paid, check them, and plug the “errors” into a formula to get the right answer. First, suppose A pays $80. That means that B pays $90 and C pays $170, giving a total of $340. That’s 500 – 340 = $160 short of the goal, so our guess of $80 yields an “error” of $160. As a second guess, suppose that A pays $150. In that case B pays $160, C pays $310, and the total is now $620. This time the “error” is 500 – 620 = -$120. The false position method (technically here the double false position method) offers this formula for finding the right answer:

In this case it gives

When A pays $120 then B pays $130, C pays 250, and together they pay $500, so this solution works.

This is hardly the most efficient way to solve a simple linear equation given the tools we have today, but it served for centuries. In his Ground of Artes of 1542, Robert Recorde offered a rule:

Gesse at this woorke as happe doth leade.
By chaunce to truthe you may procede.
And firste woorke by the question,
Although no truthe therein be don.
Suche falsehode is so good a grounde,
That truth by it will soone be founde.
From many bate to many mo,
From to fewe take to fewe also.
With to much ioyne to fewe againe,
To to fewe adde to manye plaine.
In crossewaies multiplye contrary kinde,
All truthe by falsehode for to fynde.

From Lewis Carroll’s diary, Feb. 5, 1856:

Varied the lesson at the school with a story, introducing a number of sums to be worked out. I also worked for them the puzzle of writing the answer to an addition sum, when only one of the five rows have been written: this … astonished them not a little.

He had started by writing an arbitrary number:

21879

Then he asked the students to call out a second five-digit number. Carroll added a third, the students shouted a fourth, and Carroll added a fifth and immediately wrote the sum:

  21879
  62593
  37406
  82527
+ 17472

 221877

How did he do this?

Last August, researchers at Rome University produced tiny portraits of Albert Einstein and Charles Darwin by modifying E. coli cells to respond to light patterns. Bacteria that received more light would swim faster, so over time they tended to concentrate in the darker parts of a negative image.

Lead author Giacomo Frangipane said in a statement, “Much like pedestrians who slow down their walking speed when they encounter a crowd, or cars that are stuck in traffic, swimming bacteria will spend more time in slower regions than in faster ones.”

Using the same technique, they created a (tiny) version of the Mona Lisa.

Suppress each of the digits in 1729404 and add the resulting numbers, and you get 1729404 again:

  729404
  129404
  179404
  172404
  172904
  172944
+ 172940
 1729404

Another example:

  3758846
  1758846
  1358846
  1378846
  1375846
  1375846
  1375886
+ 1375884
 13758846

There are 41 such numbers.

33 = 8866128975287528³ + (-8778405442862239)³ + (-2736111468807040)³

That result was discovered by Andrew Booker of the University of Bristol just this year.

It leaves 42 as the only positive integer less than 100 that has not been represented as the sum of three cubes.

(We can omit numbers that give a remainder of 4 or 5 when divided by 9, since those are known to be ineligible. But can every other integer be expressed in this way? It’s an open problem.)

(Thanks, Kate.)

09/06/2019 UPDATE: The case of 42 has now been solved, by Andrew Booker at Bristol and Andrew Sutherland at MIT:

42 = (-80538738812075974)³ + 80435758145817515³ + 12602123297335631³

The lowest unsolved case is now 114.