The use of historical problems in the middle school has many advantages. The dissonance created by such problems challenges the student to focus more carefully on the critical variables. These variables may be hidden in an unfamiliar context, in a language that seems different, or in a cultural setting unlike their own. Once the students have identified the critical variables they can then bring to bear their mathematical skills.

The problems that follow can be solved using the arithmetic and algebra skills typically learned in the middle school.

The use of historical problems in the middle school has many advantages. The dissonance created by such problems challenges the student to focus more carefully on the critical variables. These variables may be hidden in an unfamiliar context, in a language that seems different, or in a cultural setting unlike their own. Once the students have identified the critical variables they can then bring to bear their mathematical skills.

The problems that follow can be solved using the arithmetic and algebra skills typically learned in the middle school.

Here are some problems used by Karen Dee Michalowicz of the Langley School. She reports that her students have enjoyed solving these and have created similar problems of their own. The problems that follow come from the 1807 London Edition of John Bonnycastle’s *Introduction to Mensuration and Practical Geometry*. Two problems are in rhyme. The third problem is one that would never be printed in a book today, because of its “political incorrectness.” [*Editors' note:* By "political incorrectness," we assume the authors mean the problem might be seen to encourage underage and/or irresponsible drinking.]

In the midst of a meadow well stored with grass,

I took just an acre to tether my ass:

How long must the cord be, that feeding all round,

He may’nt graze less or more than [his] acre of ground.

Click here to reveal the answer

The answer is: 39.25073 ft

One ev’ning I chanc’d with a tinker to sit,

Whose tongue ran a great deal too fast for his wit:

He talked of his art with abundance of mettle;

So I ask’d him to make me a flat-bottom’d kettle.

Let the top and the bottom diameters be,

In just such proportion as five is to three:

Twelve inches the depth I propos’d, and no more;

And to hold in ale gallons seven less than a score.

He promis’d to do it, and straight to work went;

But when had done it he found it too scant.

He alter’d it then, but too big he had made it;

For though it held right, the diameters fail’d it;

Thus making if often too big and too little,

The tinker at last had quite spoiled his kettle;

But declares he will bring his said promise to pass,

Or else that he’ll spoil every ounce of his brass.

Now to keep him from ruin, I pray find him out

The diameter’s length, for he’ll never do’t, I doubt.

Answer: Bottom 14.44401, top 24.4002

Two porters agreed to drink off a quart of strong beer between them, at two pulls, or a draught each; now the first having given it a black eye, as it is called, or drank till the surface of the liquor touched the opposite edge of the bottom, gave the remaining part of it to the other: what was the difference of their shares, supposing the pot was the frustum of a cone, the depth being 5.7 inches, the diameter at the top 3.7 inches, and that of the bottom 4.23 inches?

Answer: 7.05 cubic inches.

The following problems come from Robert McGee of Cabrini College. In each case, he presents the problem and then gives a solution found by the students, in their own words.

The first group of problems come from various times and places:

1. A bamboo shoot 10 feet tall has a break near the top. The top touches the ground 3 feet from the stem. What is the length of the stem left standing erect ? (Yang Hui, China, 1261)

Solution:

Facts: A^{2} + B^{2} = C^{2}

A^{2} + 9 = C^{2}

C + A = 10

Try and Check

Single digits didn’t work ( ex. 4^{2} + 3^{2} is not equal to 6^{2})

Decimals with tenths didn’t work (4.6^{2} + 3^{2} is not equal to 5.4^{2})

Decimals with hundredths did work .

Answer: 4.55^{2} + 3^{2} = 5.45^{2}

4.55 is the length.

2. Ottaviano Sempioni bought three jewels, the second of which cost 4 coins more than the first, the third cost 5 coins more than the first and second together, and all three cost 81 coins. Required the cost of each. (Onofrio, *Arithmetica,* Italy, 1670)

Solution:

First I drew 3 objects representing each jewel. Then I wrote the equation for each jewel, the first being A, the second A + 4 and the third 2A + 9. Then I put the equations together making 4A + 13 = 81. Solving, A = 17, so I went back to each equation and I found my answer. 1st = 17, 2nd = 21, and 3rd = 43.

3. A hundred bushels of grain are distributed among 100 persons in such a way that each man receives 3 bushels, each woman 2 bushels, and each child half a bushel. How many men, women and children are there? (Alcuin of York, England, 8th century)

Solution. Method was guess and test. Here the first three numbers are the number of men, women, and children, respectively, while the last number is the total number of bushels.

12, 14, 74 101 12, 12, 76 98

11, 13, 76 97 13, 11, 76 99

10, 14, 76 96 14, 10, 76 100

Therefore: Men = 14

Women = 10

Children = 76

A second solution of 11 men, 15 women and 74 children was also found.

4. A powerful, unvanquished, excellent black snake, which is 80 *angulas* in length, enters into a hole at the rate of 7 1/2 *angulas* in 5/14 of a day, and in the course of 1/4 of a day, its tail grows 11/4 of an *angula*. O ornament of arithmeticians, tell me by what time the serpent fully enters into the hole. (Mahavira, India, ca. 850)

Solution:

5/14 day loses 7 1/2 *angulas*, then (5/14)(14/5) = 1 day, loses (7 1/2)(14/5) = 21 *angulas*.

1/4 day gains 11/4 *angulas*, (1/4)(4) = 1 day, gains (11/4)(4) = 11 *angulas*.

–21 + 11 = –10 *angulas* per day.

X days = –80 *angulas*, 1 day = –10 *angulas*,

8 days = –80 *angulas*

Therefore in 8 days the snake is in the hole.

5. Divide 60 into four Such parts, that the first being increased by 4, the second decreased by 4, the third multiplied 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. (Benjamin Banneker, U.S., 18th century)

Solution:

Let the numbers be X, Y, A, and B. Then

X + 4 = Y – 4 or X + 8 = Y. Also

X + 4 = 4A or (1/4)X + 1 = A. Also

X + 4 = B/4 or 4X + 16 = B. Then

X + X + 8 + (1/4)X + 1 + 4X + 16 = 60

(6 1/4)X + 25 = 60

(6 1/4)X = 35 and X = 5 3/5

Solution:

X = 5 3/5

Y = 13 3/5

A = 2 2/5

B = 38 2/5

The second group of problems is from old American textbooks.

1. If a barrel of ale will last a family of 6 persons for 2 months, how many persons would drink 9 barrels in a year?

Solution:

First state this: 1 barrel per 6 people per two months. If you divide 1 by 6 and by 2 you get .0833333...Then you state that nine barrels per X people per 12 months, or 9 divided by 12 equals .75. Then divide .75 by .0833333. By doing this you get 9 persons, which is the answer.

2. If by paying a teacher $3.25 per week a certain school would be kept 11 1/2 weeks, how long might it be kept if she be paid only $2.875 per week? (Question: When would a teacher have been earning somewhere near $3 per week?)

Solution:

First I multiplied $3.25 which she was paid by 11 1/2 weeks, which gave me $37.75. Then I divided that by $2.875 to see the number of weeks, which was 13.

3. A teacher receives 77 pieces of money for his month’s salary. The number of half dimes (nickels) received is 85 5/7% of the number of dimes; the number of dimes equals 87 1/2% of the number of quarters; the number of quarters equals 40% of the number of halves; and the number of halves equals 55 5/9 % of the number of dollars; what was his salary?

Solution:

First I converted the percents to fractions.

Nickels = 6/7 dimes

Dimes = 7/8 quarters

Quarters = 2/5 halves

Halves = 5/9 dollars

Total 77 pieces.

I first guessed 19 dollars and did try and check. My answer was way off because I needed more pieces of money. I tried higher numbers of dollars and finally got the correct answer. I guessed 36 dollars and multiplied that by 5/9. I got 20 which was the number of halves. I multiplied 20 by 2/5 and got 8 which was the number of quarters. I multiplied 8 by 7/8 and got 7 which was the number of dimes. Then I multiplied 7 by 6/7 and got 6 which was the number of nickels. I added all of the pieces of money together and got 77. The salary was $49.00. (Note how much higher this was than in the previous problem, where the teacher made somewhere near $12 for a month. Does this have to do with the fact that here the teacher is male, while in the previous problem the teacher is female?)

4. A man bequeathed $900 to three friends; the first must have a certain portion, the second must have twice as much as the first, the third $28 more than the first. How much did each person receive?

Solution:

X + 2X + (X + 28) = 900

4X + 28 = 900

4X = 872

X = $218

Thus: First gets $218

Second gets 2 x 218 or $436

Third gets $218 + $28 = $246

5. If 1 pig is worth $3 and 5 pigs are worth 2 sheep and 5 sheep are worth 2 cows and 10 cows are worth 3 horses, what is the value of 12 horses?

Solution:

First state the problem simply: 1 pig = $3, 5 pigs = 2 sheep, 5 sheep = 2 cows and 10 cows = 3 horses. 12 horses ?

I then calculated: 1 x 3 = 3; 3 x 5 = 15; 15/2 = 7.5; 7.5 x 5 = 37.5; 37.5/2 = 18.75; 18.75 x 10 = 187.5; 187.5/3 = 62.5; 62.5 x 12 = $750, the answer.

The third group of problems was designed to take advantage of the unique nature of the particular middle school where these were used, namely, the Armenian Sister’s Academy, Radnor, Pennsylvania. In addition to the standard curriculum, all students study the Armenian language and culture. Thus, these problems come from a nineteenth century Armenian arithmetic textbook.

The students first had to translate the problem from Armenian and then solve the problem. The translation from the Armenian was originally conceived of as a family project. Students were encouraged to involve their parents and grandparents in the translation. It turned out that the students were able to do a lot of the translation on their own.

The following summary describes our good fortune in discovering an Armenian arithmetic textbook that was both relevant and meaningful. The 19th century Armenian textbook, *Tvapanootyoon* (*Arithmetic*), published in 1887 in what was at the time Constantinople, Turkey, and is today Istanbul, was written to teach primary and secondary students the basics of numbers. Written by mathematician Hagop Bohajian, the book is written to accommodate the needs of beginner students, while it is graduated to challenge intermediate and advanced students as well. In an introduction to his book, Mr. Bohajian described it as comprehensive yet concise, covering a full range of computational principles and formulas. This book belonged to Mikael Markaridian, the great-grandfather of a student at the school, who resided in Turkey in the former Armenian state of Garin (Erzerum) at the time of his youth. Academy student Katrina Selverian discovered the book in her grandfather’s basement.

The book offers theoretical prose and textual instructions to guide the performance of mathematical functions, while providing problem solving sections to encourage practice and learning. Answers are printed in a final section. Mr. Bohajian warned teachers to be careful not to allow students to hasten forward through the book if they had not mastered the previous sections. He assured them, simultaneously, that any student who mastered the entire book would be able to accomplish any mathematical task encountered in business with facility and ease.

The problems here are presented in the English translation by the students.

1. A leaves a city at 9:15 and goes to another city going 4 mph. B leaves this second city at 9:30 and goes to the first city going 3 1/2 mph. The 2 cities are 21 miles apart. At what time are A and B going to meet?

Solution:

At 9:15 A is moving 4mph and B is not moving yet. We then found that at 9:30, the two people were 20 miles apart. At 10:30 they were 12 1/2 miles apart. We added 4 and 3 1/2 and got 7 1/2. It was 20 miles at 9:30, which is 7 1/2. So 20 mi / 7 1/2 mph = 2 2/3 hours = 2 hr. and 40 min. after 9:30, which equals 12:10. Therefore A and B will meet at 12:10.

2. There is a garden with fruit trees, 1/2 of them are apple trees, 1/4 of them are peach trees, and 1/6 of them are plum trees. The remaining trees are 200 cherry trees. How many trees are there in the garden?

Solution:

We wrote out all of the fractions 1/2, 1/4, and 1/6 and found a common denominator which was 12. We then added up all the fractions to get 11/12. We saw that if you added 1/12, you would get 1. 1/12 is the number of cherry trees which is 200. Multiply 200 by 12 which equals 2400 which is the number of trees in the garden.

3. A man who sells eggs, sells half his eggs and one more to the first buyer. To the second buyer he sells half of the remaining eggs and one more. The third buyer buys half of the remaining eggs and one more, and there are no more eggs left. How many eggs were there in the beginning?

Solution:

I did this problem by working backwards.

(((((0 + 1) x 2) + 1) x 2) + 1) x 2 = 14, the number of eggs he started with.

4. If 30 men need 40 days to complete a job, how many men will it take to complete a job that is 5 times bigger in one-fifth the amount of time?

Solution:

There are 30 men, 40 days, and 1 project, (30,40,1). Since the project is 5 times as big, it becomes (30,40,5). Since it must take 1/5 of 40 days or 8 days, it then becomes (30,8,5). You multiply 5 (from the 1/5) by 5 ( the size of the new project) , which equals 25. You then multiply 25 by 30 (men) which equals the answer. 750 men are needed to finish 5 times the project in 8 days.

5. A man left 38,000 dollars that was to be divided among his three sons and his three daughters. Each son is going to get 33 1/3 % more than the oldest daughter; and the younger daughters are going to get 33 1/3 % less than the oldest daughter. What is everyone’s portion?

Solution:

Let S = son, D = youngest daughter, X = oldest daughter. Then S = (1/3)X + X and D = X – (1/3)X. The equation for the entire amount is 3S + 2D + X = 38,000. Substituting, 3((1/3)X + X) + 2(X–(1/3)X) + X = 38,000. Simplifying we got (6 1/3)X = 38,000. X = 6000 which is what the oldest daughter gets. We substituted and got S = 8000 and D = 4000.

A final problem comes from medieval Europe. Alcuin of York (c. 735-804), the educational advisor to Charlemagne, left a collection of mathematics problems in a work entitled *Propositions for Sharpening Youthful Minds*. Among the problems was the following:

A dying man left 960 shillings and a pregnant wife. He directed that if a son was born he should receive 9/12 of the estate and the mother should receive 3/12. If however a daughter was born, she should receive 7/12 of the estate and the mother should receive 5/12. It happened however that twins were born, a boy and a girl. How much should the mother receive, how much the son, how much the daughter?

This problem turns out to be a particularly “rich” one. I gave this problem to groups of my 8th grade students as part of a problem-solving project using historical problems. The solutions given are those that the students found.

One group came up with the following solution.

We came to our conclusion because we knew that the boy receives 2/12 more than the girl and the girl receives 2/12 more than the mother. We then split up the fractions and figured out that the mother receives 2/12, the girl 4/12 and the boy 6/12, and it all adds up to 12/12 or 960 shillings, Boy - 480, girl - 320, mother - 160.

A second group arrived at a different solution.

The boy would get 9/12 of 960 or 75%. The girl would get 7/12 or 58%. You take the mother’s 25% and 42% and add them and divide them by two. If there were twins you do this to find out what the mother’s share is. Her share is 1/3 or 320 out of 960. 960 - 320 = 640. 133% is the sum of the boy’s and girl’s share. To find out what the boy would get, you divide 640 by 133% . Then multiply by 75%. The son gets 360. The daughter gets 280. The mom gets 320.

This is essentially the solution given by Alcuin of York. It is interesting to note the appeal of this logic. I attended a workshop at a NCTM Regional Meeting in which about 25 participants were divided into cooperative learning groups. All groups came up with the above solution, with no dissenters.

One student arrived at the following solution.

S + D + M = 960.

S = 3M D = 7/5M.

3M + 7/5 M + M = 27/5M = 960. Therefore M = 177 7/9.

Son gets 533 3/9. Daughter gets 248 8/9. Mother gets 177 7/9.

This solution uses the same logic as Nicolas Chuquet, who presented a similar problem in his *Triparty* of 1484:

A man makes a will and dies leaving his wife pregnant. His will disposes of 100

ecussuch that if his wife has a daughter, the mother should take twice as much as the daughter, but if she has a son, he should have twice as much as the mother. The mother gives birth to twins, a son and a daughter. How should the estate be split, respecting the father's intentions?

Chuquet's solution is that the inheritance should be divided in the ratio of 4:2:1.

As a postscript, when a related problem appeared in the column, "Ask Marilyn" in *Parade Magazine*, 10/05/03, the response given was “The estate should be divided according to the law. If the conditions of the will cannot be met, it is invalid. Without a will, the widow would receive half or a third of the man’s estate, and the children would share the remainder.” Interestingly enough, Robert Recorde, in his *Grounde of* *Artes* (c. 1542), wrote, "If some cunning lawyers had this matter in scanning, they would determine this testament to be quite voyde, and so the man to die untestate, because the testament was made unsufficient." Is this thirteen hundred years of progress in solving this problem?

More on this problem may be found in D. E. Smith, *History of Mathematics*, vol. II, pp. 544-546 (New York: Dover, 1958) and in Singmaster and Hadley, "Problems to sharpen the young," *Mathematical Gazette, *No. 475 (March, 1992), 102-126. The nineteenth century arithmetic text by Edward Brooks, *The Normal Written Arithmetic* (Sower, Barnes and Potts, 1863), contains similar problems.