Rudolff's problem
Problem
Rudolff's book, called Coss, written in 1525, was the first German algebra book. The reason for the title is that cosa is a 'thing', a term used for the unknown. Algebraists were called cossists, and algebra the cossic art, for many years. Here is the problem:
A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £ 2 for women and 50p for children. How many men, women and children are there in the group?
Student Solutions
Congratulations to Elizabeth Whitmore and Sue Liu of Madras College, St Andrew's, Scotland, to Giulio Tiozzo, age 15, Liceo Scientifico "Galileo Ferraris", Turin, Italy, and to Vassil Vassilev, age 14, Lawnswood High School, Leeds and Jonathan Kemp, age 17, Westwood High School, Leek, England. Here is Jonathan's solution.
The problem states that a group of 20 people pay a total of £20 to see an exhibition, admission £3 for men, £2 for women and 50p for children.
Denoting the number of men as $x$, the number of women as $y$ and the number of children$z$ you have to solve two simultaneous equations:$$ x + y + z = 20$$ and $$3x + 2y + {z\over 2} = 20.$$ Multiplying the second equation by 2 and subtracting the first equation to eliminate $z$ gives: $$ 5x + 3y =20.$$ This is a Diophantine equation as featured in this months article. It is easy to spot two solutions, namely $x = 4$ , $y = 0$, and $x = 1$, $y = 5$. In this case we have 2 simultaneous equations and 3 unknowns.
Substituting these solutions into the first two equations gives: $ 4 + 0 + z = 20$ therefore $z = 16$ and $12 + 0 + z/2 =20$ and again $z = 16$.
Substituting $x = 1$ and $y = 5$ gives: $1 + 5 + z = 20$ therefore $z = 14$ and $3 + 10 + z/2 = 20$ and again $z = 14$.
So the group can either consist of either 4 men, no women and 16 children, or of 1 man, 5 women and 14 children.