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 ecus such 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.