The Mathematical Puzzles of Benjamin Banneker
Introduction
In his journals, Benjamin Banneker (a self-taught mathematician and scientist) wrote and collected mathematical puzzles written in verse. The mathematics in this journal consisted of six puzzles and two pages of mathematical writing. Banneker’s six puzzles were published in an excellent biography of him written by Silvio Bedini. To my knowledge, I am able to reproduce here for the first time, three of Banneker’s puzzles in his actual handwriting.^{i}
Using Banneker’s Mathematical Puzzles
Benjamin Banneker’s puzzles can be solved by middle and high school students. Banneker didn’t use symbolic algebra to solve these problems, and they illustrate that some problems are most easily solved without algebra! The mathematics department at my high school sponsored a contest centered on these puzzles. Each grade was assigned a problem to solve, and one problem was open to all students. The students had a number of weeks to solve the problems, and savings bonds were awarded to randomly chosen correct entries. We assigned the problems to the four grades as follows:
Ninth Grade | Puzzle #3: “Divide 60 into four such parts...” |
Tenth Grade | Puzzle #1: “A gentleman Sent his Servant...” |
Eleventh Grade | Puzzle #2: “Suppose a ladder 60 feet long...” |
Twelfth Grade | Puzzle #4: “A, B, and C, discoursing about their ages...” |
All Grades | Puzzle #6: “The Puzzle of the Hound and the Hare” |
Banneker's Mathematical Methods
Banneker used the single and double position methods, which are early techniques for solving problems that are difficult to translate into equations. The following definition comes from Daniel Adams’s 1802 book The Scholar’s Arithmetic or Federal Accountant:
"Position is a rule which, by false or supposed numbers, taken at pleasure, discovers the true one required. It is of two kinds, Single and Double. Single Position is the working with one supposed number, as if it were the true one, to find the true number.
Rule:
- Take any number and perform the same operations with it as are described in the question.
- Then say, as the sum of the errors is to the given sum, so is the supposed number to the true one required.
Proof
- Add the several parts of the sum together, and if it agrees with the sum, it is right."
This method assumes that the expression is close to or exactly linear. Using the above rule, call x_{1} the guess (the supposed number) and y_{1} the result of applying the rule to x_{1}. If y_{2} is the "given sum," then the rule dictates that:
This is true if the expression is linear and almost true if it is approximately linear.
Translated, this method involves a first guess at the solution to a problem. If that guess is not correct, then the guess should be multiplied by the ratio of the desired result to that incorrect result, called the "correction factor" by Professor Lumpkin (1996), to get the correct solution.
Double Position involves the use of two guesses to find the value, which makes an expression equal to zero. After Rule 1, Adams states, "2. Place each error against its respective position or supposed number; if the error be too great, mark it with a +; if too small with a -. 3. Multiply them cross-wise, the first position by the last error, and the last position by the first error. 4. If they be alike, that is, both greater or both less than the given number, divide the difference of the products by the difference of the errors, and the quotient will be the answer; but if the errors be unlike, divide the sum of the products by the sum of the errors, and the quotient will be the answer."
Again, it is based on the assumption that the expression is either linear or approximately linear. Essentially the method calculates the x-intercept of the line passing through the two guessed points. Suppose guess x_{1} produces a result of y_{1} and that guess x_{2} produces a result of y_{2}. Let x be the "true" answer that makes the expression equal to 0. Then by setting slopes equal, we get:
For example, with g(x) = x^{2} - 5, let (x_{1},y_{1}) = (3, 4) and let (x_{2}, y_{2}) = (1, -4). Then:
If (x_{2}, y_{2}) is changed to (2, -1), then the estimate for the x-intercept would be:
The correct answer is 2.2361.
Puzzle 1
A gentleman Sent his Servant with £100 to buy 100 Cattle, with orders to give £5 for each Bullock, 20 Shillings for cows, and one Shilling for each Sheep, the question is to know what number of each sort he brought to his master.
Aswer
19 Bullocks at 5£ each | £95 |
1 Cow at 20 S. | 1 |
80 Sheep at 1S each | 4 |
100 proof | 100 |
Note: A shilling equals 1/20 of a pound (£).
Answer: Using the obvious variables, we get: . This set of Diophantine equations has only one set of positive integer solutions: B = 19, C = 1, and S = 80.
Puzzle 2
Suppose ladder 60 feet long be placed in a Street so as to reach a window on one Side 37 feet high, and without moving it at bottom, will reach another window on the other side of the Street which is 23 feet high, required the breadth of the Street.
The bottom of a 60-foot ladder is placed on a street. If the top is swung to one side of the street, it reaches a window 37 feet high. If the top, without moving the bottom, is swung to the other side of the street, it reaches a window that is 23 feet high. How wide is the street?
Answer: Using the Pythagorean theorem, we get the expressions shown in this screen shot:
And therefore the width of the street is 102.65 feet.
Puzzle 3
Question by Ellicott Geographer General
Divide 60 into four Such parts that the first being increased by 4, the Second decreased by 4, the third multiplyed by 4, the fourth part divided by 4, that the Sum, the difference, the product, and the Quotient shall be one and the Same number.
Ans | ||
first part 5.6 increased by 4 | 9.6 | |
Second part 13.6 decreased by 4 {is} | 9.6 | |
third part 2.4 multiplyed by 4 | 9.6 | |
fourth part 38.4 divided by 4 | 9.6 | |
60.0 |
Answer: Using the Single Position method, guess at the "same number." Let's guess 16.
Then the first part is 12 (12 + 4 = 16),
the second part is 20 (20 - 4 = 16),
the third part is 4 (4 x 4 = 16),
and the fourth part is 64 (64 / 4 = 16).
These four parts add up to 12 + 20 + 4 + 64 = 100.
Therefore, the correction factor is 60/100, or 3/5.
Thus the answer is the guess, 16, times the correction factor, 3/5; 9.6 is the value for the "same number." The four parts, then, are 5.6, 13.6, 2.4, and 38.4—their sum to 60, the desired result.
Single Position works here because, since the guess produced too large a result, it is reduced by a correction factor based on the ratio of the desired answer to the incorrect answer.
Puzzle 4
A, B, and C, discoursing about their ages, Says A, if from double the cube root of B’s age, double the biquadrate root of C’s age betaken the remainder will be equal to the Sursolid root of my age, Says B, the Square root of my age is equal to one fourth part of A’s, and Says C, the Square root of my age is one more than the Square root of B’s, Required their Several ages.
A's | 32 The Sursolid root of which is 2 | |
B's | {Age} | 64 The Cube root of which is 4 |
C's | 81 The biquadrate root of which is 3. |
Notes: "Biquadrate" means square of the square. "Sursolid" means fifth power.
In modern language:
Three people—call them A, B, and C—talk about their ages.
A states: If I subtract twice the fourth root of C’s age from twice the cube root of B’s age, I get the fifth root of my age.
B states: The square root of my age equals one-fourth of A’s age.
C states: The square root of my age is one greater than the square root of B's age.
How old are A, B, and C?
Answer: The equations that follow are: .
Banneker correctly assumed that (1) each age was an integer, and (2) since the fifth root of A’s age was used, the only reasonable value of A was 32 (Lumpkin 1996). This gives a value of 64 from the second equation for B, and thus a value of 81 from the third equation for C. Testing these values in the first equation we get: .
Puzzle 5
This particular puzzle is attributed to Banneker and was originally published in a mid-nineteenth century biography of Banneker by Martha Tyson.
A cooper and vintner sat down for a talk,
Both being so groggy that neither could walk;
Says cooper to vintner, "I'm the first of my trade,
There's no kind of vessel but what I have made,
And of any shape, sir, just what you will,
And of any size, sir, from a tun to a gill."
"Then," says the vintner, "you're the man for me.
Make me a vessel, if we can agree.
The top and the bottom diameter define,
To bear that proportion as fifteen to nine,
Thirty-five inches are just what I crave,
No more and no less in the depth will I have;
Just thirty-nine gallons this vessel must hold,
Then I will reward you with silver or gold, —
Give me your promise, my honest old friend."
"I'll make it tomorrow, that you may depend!"
So, the next day, the cooper, his work to discharge,
Soon, made the new vessel, but made it too large:
He took out some staves, which made it too small,
And then cursed the vessel, the vintner, and all.
He beat on his breast, "By the powers" he swore
He never would work at his trade any more.
Now, my worthy friend, find out if you can,
The vessel's dimensions, and comfort the man!
Notes:
A "tun" is a large cask, frequently used for wine. In England, a tun contains 252 wine gallons.
"Gill": there are four gills in a pint.
The ratio of the diameters of the top and bottom are 15/9. This phrase also implies that the top and bottom are circles.
The height of the vessel is 35 inches. The capacity is 39 gallons. In the United States, a gallon is 231 in.^{3}. A British imperial gallon is 277.42 in.^{3}. In 1854, Benjamin Hallowell of Alexandria, Virginia, proposed a solution to this problem at a meeting of the Maryland Historical Society. His solution of 24.745 inches and 14.8476 inches for the diameters is based on the size of an ale gallon, which was 288 in.^{3}.
Answer: I assumed that the vessel was in the shape of a frustrum of a cone. Let h represent the height of the cut-off part of the cone; then by similar triangles, , so h = 52.5. I used British imperial gallons, which gave the volume as 39 x 277.42 = 10819.36 in.^{3}. Letting k represent the constant of proportionality of the radii of the top and bottom, the volume of the frustrum is the difference between the volumes of two cones:
Setting this difference equal to 10819.36 in.^{3} gives k = 0.81815, which yields diameters of 24.5445 and 14.7267 inches.
Puzzle 6
When fleecy skies have Cloth'd the ground
With a white mantle all around
Then with a grey hound Snowy fair
In milk white fields we Cours'd a Hare
Just in the midst of a Champaign
We set her up, away she ran,
The Hound I think was from her then
Just Thirty leaps or three times ten
Oh it was pleasant to see
How the Hare did run so timorously
But yet so very Swift that I
Did think she did not run but Fly
When the Dog was almost at her heels
She quickly turn'd, and down the fields
She ran again with full Career
And 'gain she turn'd to the place she were
At every turn she gan'd of ground
As many yards as the greyhound
Could leap at thrice, and She did make,
Just Six, if I do not mistake
Four times She Leap'd for the dogs three
But two of the Dogs leaps did agree
With three of hers, nor pray declare
How many leaps he took to Catch the Hare.
(Answer)
Just Seventy two I did Suppose,
An Answer false from thence arose,
I doubled the Sum of Seventy two,
But still I found that would not do,
I mix'd the Numbers of them both,
Which Shew'd so plain that I'll make Oath,
Eight hundred leaps the Dog did make,
And Sixty four, the Hare to take
Notes:
- Initially the dog is 30 dog leaps from the hare.
- The hare turns when the dog reaches the hare.
- The hare gains three dog leaps' worth of yards at each turn.
- The hare makes six turns.
- Three times the number of hare leaps = four times the number of dog leaps.
- Two times the length of each dog leap = three times the length of each hare leap.
- The question: How many leaps does the dog make in order to catch the hare?
- Statements (5) and (6) combine to state that the dog runs 9/8 as fast as the hare.
- Statements (3) and (4) combine to the effect that through turning, the hare gained a total of 18 dog leaps.
- Statements (1) and (9) combine to the equivalent of giving the hare a 30 + 18 = 48-dog-leap head start.
Assumption: Banneker's arithmetic at the end is equivalent to solving the proportion , which gives x = 864. This is illustrative of the method of False Position.
Answer: Banneker started with 72 as a guess and got a value of 4 rather than the desired 48 (the hare’s equivalent head start). Banneker must have calculated how far the hare would run (in dog leaps), if the dog ran 72 dog leaps, and got an answer of 68. This meant that if the hare had a head start of 4 dog leaps, the dog would catch the hare in 72 leaps. He then used the correct head start, 48, to get his answer of 864. Unfortunately, Banneker made an error. The dog runs 9/8 as fast as the hare, and therefore when the dog runs 72 dog leaps, the hare runs a distance of 64 dog leaps—leaving a difference of 8. Replacing the “4” in Banneker’s computation with 8 gives the correct answer to the puzzle of 432 dog leaps.
Another method is supported by the last eight lines, which are entitled “Answer.” The author, presumably Banneker, starts off with 72 as a guess and gets a “false answer.” Then 72 is doubled to get 144, but “that would not do.” “I mixed the Numbers of them both” is illustrative of the method of Double Position. Assuming the dog takes 72 leaps, the hare, traveling 8/9 as fast as the dog, travels the distance of 64 dog leaps. The hare also has a 48-dog-leap head start. So at the end of the 72 dog leaps, the hare is (64 + 48) - 72 = 40 dog leaps ahead of the dog. Assuming the dog takes 144 leaps, the hare, traveling 8/9 as fast as the dog, travels the distance of 128 dog leaps. The hare also has a 48-dog-leap head start. So at the end of the 144 dog leaps, the hare is (128 + 48) - 144 = 32 dog leaps ahead of the dog. In modern notation, the rule of Double Position states that . Letting (x_{1}, y_{1}) = (72, 40) and (x_{2}, y_{2}) = (144, 32), we get .
References
Adams, Daniel. Scholars Arithmetic or Federal Accountant. Leominster, Massachusetts, 1802.
Bedini, Silvio A. The Life of Benjamin Banneker: The First African American Man of Science. 2nd ed. Baltimore, Maryland: Maryland Historical Society, 1999.
Benjamin Banneker Association. http://www.mth.msu.edu/banneker/about.html.
Eglash, Ron. "The African Heritage of Benjamin Banneker." In Social Studies of Science. Vol. 27, no. 2 (April 1997).
Fasanelli, Florence D. "Benjamin Banneker's Life and Mathematics: Web of Truth? Legends As Facts; Man vs. Legend," a talk given on January 8, 2004, at the MAA/AMS meeting in Phoenix, Arizona.
Lumpkin, Beatrice. "From Egypt to Benjamin Banneker: African Origins of False Position Solutions." In Vita Mathematica, Historical Research and Integration with Teaching. MAA Notes, vol. 40 (1996).
Lumpkin, Beatrice. "Mathematical Puzzles and Exercises from Banneker's Manuscript Journal." From an unpublished manuscript. Dr. Lumpkin taught at Malcolm X College in Chicago.
Mahoney, John F. "Benjamin Banneker's Mathematical Puzzles." In NCTM's Mathematics Teacher. Vol. 96, no. 2 (February 2003).
Mahoney, John F. "Benjamin Banneker and Single Position," In NCTM's Mathematics Teaching in the Middle School. Vol. 10, no. 7 (March 2004).
Mahoney, John F. "Benjamin Banneker and the Law of Sines," an article in the review process for NCTM's Mathematics Teacher.
Tyson, Martha E. Banneker, The Afric-American Astronomer. Philadelphia Friends Book Association, 1884. (Copy at the Maryland Historical Society.)
^{i}I located a microfilmed copy of Banneker’s journal at the Maryland Historical Society in Baltimore. The quality of the reproduction was poor, but I was helped by the painstakingly accurate photographic retouching by Mr. Omar Rumi of Kuala Lumpur, Malaysia, and by my son, Quinn, a student at MIT.
Authored by
- John F. Mahoney
Benjamin Banneker Academic High School
Washington, District of Columbia