Number eight cells:

Now suppose we want to color each cell red or blue such that no three cells are in arithmetic progression — for example, we don’t want cells 1, 2, and 3 to be the same color, or 4, 6, and 8. With eight cells it’s possible to accomplish this:

But if we want to add a ninth cell we can’t avoid an arithmetic progression: If the ninth cell is blue then cells 1, 5, and 9 are evenly spaced, and if it’s red then cells 3, 6, and 9 are. Dutch mathematician B.L. van der Waerden found that there’s always such a limit: For any given positive integers r and k, there’s some number N such that if the integers {1, 2, …, N} are colored, each with one of r different colors, then there will be at least k integers in arithmetic progression whose elements are of the same color. Determining what this limit is (in this example it’s 9) is an open problem.

(Bonus: Alexej Kanel-Belov found this pretty theorem concerning divisibility of integer sums within an infinite grid — Martin J. Erickson, in Beautiful Mathematics, calls it a two-dimensional version of van der Waerden’s theorem.)

A stick is broken at random into 3 pieces. It is possible to put them together into the shape of a triangle provided the length of the longest piece is less than the sum of the other 2 pieces; that is, provided the length of the longest piece is less than half the length of the stick. But the probability that a fragment of a stick shall be half the original length of the stick is 1/2. Hence the probability that a triangle can be constructed out of the 3 pieces into which the stick is broken is 1/2.

— Samuel Isaac Jones, Mathematical Wrinkles, 1912

(The actual probability is 1/4.)

Pick a positive integer, list the positive integers that will divide it evenly, add these up, and subtract the number itself:

• 10 is evenly divisible by 10, 5, 2, and 1. (10 + 5 + 2 + 1) – 10 = 8.

Now do the same with that number, and continue:

• 8 is evenly divisible by 8, 4, 2, and 1. (8 + 4 + 2 + 1) – 8 = 7.
• 7 is evenly divisible by 7 and 1. (7 + 1) – 7 = 1.
• 1 is evenly divisible only by 1. (1) – 1 = 0.

Many of these sequences arrive at some resolution — they terminate in a constant, or an alternating pair, or some regular cycle. But it’s an open question whether all of them do this. The fate of the aliquot sequence of 276 is not known; by step 469 it’s reached 149384846598254844243905695992651412919855640, but possibly it reaches some apex and then descends again and finds some conclusion (the sequence for the number 138 reaches a peak of 179931895322 but eventually returns to 1). Do all numbers eventually reach a resolution? For now, no one knows.

Another contribution from Lee Sallows:

“The smallest, oldest and most famous magic square of all is the specimen of Chinese origin known as the Lo shu. In this, the numbers from 1 to 9 are so placed that their sum taken in any row, column or diagonal is 15. This is another way of saying that the sum of any three of them lying in a straight line is 15. Less well known is the ‘Egyptian’ Lo shu (seen below) in which the same numbers are rearranged in a triangular formation that exhibits the same property.”

(From his book Geometric Magic Squares, 2013.) (Thanks, Lee.)

A cab was involved in a hit and run accident at night. Two cab companies, the Green and the Blue, operate in the city. 85% of the cabs in the city are Green and 15% are Blue.

A witness identified the cab as Blue. The court tested the reliability of the witness under the same circumstances that existed on the night of the accident and concluded that the witness correctly identified each one of the two colors 80% of the time and failed 20% of the time.

What is the probability that the cab involved in the accident was Blue rather than Green knowing that this witness identified it as Blue?

Psychologists Amos Tversky and Daniel Kahneman offered this problem to study subjects in 1972. The right answer is about 41 percent:

• There’s a 12% chance (15% times 80%) of the witness correctly identifying a blue cab.
• There’s a 17% chance (85% times 20%) of the witness incorrectly identifying a green cab as blue.
• Thus there’s a 29% chance (12% plus 17%) that the witness will identify the cab as blue.
• And that means there’s approximately a 41% chance (12% divided by 29%) that the cab identified as blue is really blue:

Most subjects estimated the probability at more than 50 percent, some more than 80 percent.

Tversky and Kahneman call this the representativeness heuristic: When we rely on representativeness to make a judgment, we tend to judge wrongly because the fact that a thing is more representative doesn’t make it more likely.

(Amos Tversky and Daniel Kahneman, “Evidential Impact of Base Rates,” No. TR-4, Stanford University Department of Psychology, 1981.)

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.