What Is the Eureka Effect?
10 illustrative math puzzles.
Posted Jan 20, 2021
A well-known anecdote is told about the great mathematician Archimedes. While he was about to take a bath, he suddenly realized that the volume of water displaced as he sank his body into the tub was equal to the volume of the part of his submerged body. Excited by his unexpected discovery, Archimedes purportedly leaped from the tub, running naked in the streets, shouting Eureka—“I have found it (the solution).”
This story was recounted nearly 200 years after the alleged event by the Roman writer Vitruvius, which is now thought to be largely made up by him. Whether true or not, it encapsulates how solutions to non-obvious problems are discovered by happenstance and how this produces a unique kind of elation—now called the Eureka effect.
Much research has been conducted in psychology on this effect, also called the Aha effect. But anyone can experience it concretely by doing math puzzles. These require the same type of thinking exhibited by mathematicians and scientists; as such, they show that it is possessed by everyone, no matter who they are.
One of the oldest mathematical texts in existence, the so-called Rhind Papyrus, dating back to before 1650 BCE (the year in which it was copied), produces the Eureka effect, to varying degrees, to this day, with many of its over 80 problems and puzzles. It showed that mathematics was all about discovering non-obvious patterns, which are called "obscure secrets" in the preface to the papyrus.
The late Martin Gardner showcased this effect brilliantly to modern audiences, with his intriguing book, Aha! Insight! Gardner firmly believed that math puzzles were the main tools for teaching mathematics. It is to be noted that the Rhind Papyrus was designed as a school text for Egyptian students of mathematics.
The 10 puzzles I have selected will (presumably) trigger the Eureka effect. I have adapted most of them from classic puzzle collections, modifying them for the sake of convenience and illustration.
What makes these puzzles especially intriguing for me is that they never lose their appeal or their mystery. Each time I come back to them after a while, I have to figure them out anew, as if I had never solved them in the first place. I am not sure why this is so. One thing is for certain: They show in their miniature way that discovery starts by considering what is non-obvious. As the Scottish mathematician, Eric Temple Bell, once wrote: “Obvious is the most dangerous word in mathematics.”
1. Let’s start with a puzzle adapted from the Rhind Papyrus.
Below is an estate inventory:
sheaves of wheat 2,401
hekats of grain 16,807
The first five figures represent consecutive powers of seven: 7 = 71, 49 = 72, 343 = 73, 2,401 = 74, and 16,807 = 75. But the last figure, 19,607, does not. How does it fit in?
2. A tale that is told about the German mathematician Karl Friedrich Gauss, who was only nine years old when he purportedly dazzled his teacher, a certain J. G. Büttner, with his exceptional mathematical abilities. One day, his class was asked to cast the sum of all the numbers from one to one hundred: 1 + 2 + 3 + 4 + … + 100 = ? Gauss raised his hand within seconds, giving the correct response of 5,050, astounding both his teacher and the other students who continued to toil over the gargantuan arithmetical task before them. How did Gauss do it? Although this is not a puzzle in the normal sense, it makes evident how the “Eureka mind” operates.
3. This puzzle comes from the classic collection, Cyclopedia of 5000 Puzzles, by Sam Loyd (1841-1911), one of the greatest puzzle makers of all time, and one of my own favorites. I have simplified it considerably from the original, (hopefully) retaining its Eureka form.
At an archery competition, a certain young woman carried off the first prize. Each of her separate scores on the target were either 16 points or 17 points. Altogether, the scores added up to 100 points. Can you figure out how many arrows she must have used to accomplish the feat?
4. This puzzle is almost the same to the estate puzzle above. But it appears as a popular seventeenth century popular English nursery rhyme. I put it here because of this incredible coincidence, and leave it at that. By the way, there is a trick in the wording of the puzzle.
As I was going to St. Ives
I met a man with seven wives.
Each wife had seven sacks,
Each sack had seven cats,
Each cat had seven kits.
Kits, cats, sacks, wives,
How many were going to St. Ives?
5. This is one of the best-known puzzles of all time. I have used constantly in writing about the nature of puzzles because it shows how some puzzles use traps to lead us astray in our search for the answer. I am not sure what its original source is. The particular version below (adapted slightly) turned up in an arithmetic textbook written by Christoff Rudolf and published in Nuremberg in 1561. But there are antecedents. The same type of puzzle, with different details, is found in the third section of Fibonacci’s Liber Abaci of 1202.
A snail is at the bottom of a 30-foot well. Each day it crawls up 3 feet and slips back 2 feet. The snail has the ability to stick to the walls of the well and, thus, does not slide down to the bottom at the end of a day when it stops to rest. At that rate, on what day will it reach the top of the well and crawl out?
6. I am not sure what the source of this next puzzle is. Looking around, I came across some mention that it emerged at MIT in the early 1940s, although I could not confirm this. It became popular as an item in puzzle books starting in the 1960s.
A census taker asks a resident about her children. The resident says, “I have three children. The product of their ages is thirty-six. The sum of their ages is the number on the gate to my house.” The census taker says that this is not enough information. However, after the woman says, “My oldest child has the measles,” the census taker figures it out. What’s are the children’s ages?
7. Below is a famous age puzzle, not invented by the ancient mathematician Diophantus, despite the fact that it is about him. It was discovered in an early medieval compilation, called the Greek Anthology, which historians credit to the Greek poet, mathematician, and grammarian Metrodorus (c. 500 CE). Interestingly, a number of the puzzles in the anthology are identical to those found in the Rhind Papyrus:
Diophantus’ boyhood lasted 1/6 of his life; he married after 1/7 more; his beard grew after 1/12 more, and his son was born 5 years later; the son lived to half his father’s age, and the father died 4 years after the son. How old was Diophantus when he died?
8. This next puzzle is found in all kinds of math textbooks and puzzle anthologies. The reason may be that it shows how we sometimes do not see the obvious. It is thus a kind of reverse-psychology puzzle, in a manner of speaking.
What two whole numbers (not fractions) make the number 23 when multiplied together?
9. This next puzzle gets us to reflect on the relation between symbols and math, which is what semioticians (like myself) study all the time. It is found in all kinds of math textbooks and puzzle books.
What math symbol can be put between 5 and 7 to produce a number that is greater than 5 but smaller than 7?
10. The great British puzzle-maker, Henry E. Dudeney (1857-1930) created a host of math puzzle genres. One of these involves figuring out family relations. The one below, which is not by Dudeney, falls under this rubric. I was not able to find its congener; but it is everywhere in math textbooks and puzzle collections.
Hungry Horace was thumbing through his family photograph album, which has a photo of his parents, grandparents, all the way up to each of his great-great-great-grandparents. How many photos is that?
1. It represents the sum of these numbers: 7 + 49 + 343 + 2,401 + 16,807 = 19,607. It is thus the total value of the estate.
2. In the sequence of numbers from 1 to 100, by adding the first number to the second last, the second to the third last, and so on, we will get 49 pairs that add up to one hundred: 1 + 99 = 100, 2 + 98 = 100, 3 + 97 = 100, and so on up to 49 + 51 = 100. That makes 4,900. The number 50, being in the middle, stands alone, as does 100, standing at the end. Adding 50 and 100 to 4,900 gives 5,050.
3. Six arrows, given that the only addition of the score points adding up to 100 is as follows: 17 + 17 + 17 + 17 + 16 + 16 = 100.
4. The puzzle asks how many kits, cats, sacks, and wives were going to St. Ives. None were going to it; they were all coming from it. Only one person was going to St. Ives—the narrator of the rhyme.
5. Since the snail crawls up 3 feet, but slips back 2 feet, its net distance gain at the end of every day is 1 foot up from the day before. To put it another way, the snail’s climbing rate is 1 foot up per day. At the end of the first day, therefore, the snail will have gone up 1 foot from the bottom of the well, and will have 29 feet left to go to the top on the second day (remembering that the well is 30 feet in depth).
Day 1: Goes up to the 3-foot mark and slides down to the 1-foot mark.
Day 2: Starts at the 1-foot mark, goes up to the 4-foot mark and slides down to the 2-foot mark.
Day 3: Starts at the 2-foot mark, goes up to the 6-foot mark and slides down to the 3-foot mark.
Day 26: Starts at the 25-foot mark, goes up to the 28-foot mark and slides down to the 26-foot mark.
Day 27: Starts at the 26-foot mark, goes up to the 29-foot mark and slides down to the 27-foot mark.
Day 28: Starts at the 27-foot mark, goes up 3 feet, reaches the top, and goes out, precluding its slippage back down. So, the snail gets to the top on the 28th day..
6. There are various possibilities for three numbers standing for the children’s three ages, which, when multiplied together, must give 36. But only 9, 2, 2, and 6, 6, 1 add up to the same sum, which is the number of the gate: 9 × 2 × 2 = 36 and 6 × 6 × 1 = 36; and 9 + 2 + 2 = 13 and 6 + 6 + 1 = 13. No other triplet of numbers will satisfy these two conditions; that is, there is no other set of three numbers that will produce a product of 36 and the sum of 13. When the resident stated that her “oldest” child had the measles, the census taker realized that the triplet that allowed for one, and only one, oldest child was 9, 2, 2. Therefore, the ages of the children are 9, 2, and 2. In 6, 6, 1 there are two, not one, “older” children.
7. By letting Diophantus’ age be x, the following equation translates the statement into appropriate algebraic language: x/6 + x/7 + x/12 + 5 + x/2 + 4 = x. Solving for x we get 84.
8. It is easy to get duped by this puzzle. The only possible answer is: 23 x 1 = 23.
9. A decimal point will do the trick: 5.7
10. This is a puzzle based on the exponential number 2n.
2 + 4 + 8 + 16 + 32 = 62 photos in all.