#2           57. TO MARKET, TO MARKET                  

In the 18th century there was published in England a popular periodical called The Ladies’ Diary or The Women’s Almanack, containing many ‘Delightful and Entertaining Particulars Peculiarly Adapted for the Use and Diversion of the Fair Sex’.

One such ‘Peculiar Diversion’ deals with three Dutchmen and their wives who went marketing for pigs.

The men’s names were Henry, Ely and Cornelius; the women’s Gertrude, Catherine and Anna.

Each person bought as many pigs as he or she paid guilders for each pig, and each man spent 63 guilders more than his wife.

Henry bought 23 more pigs than Catherine, and Ely 11 more than Gertrude.

What was the name of each man’s wife?

 Hints and strategies 

Hint 1 

Hint 2 

Solution

Extensions

HINT 1

The problem here is how best to assemble the given information.

As a suggestion, if M denotes a man and W his wife, write down an expression involving M and N.

HINT 2

Did you get

M2W2 = 63?

What sort of numbers are M and W? And what are the factors of 63?

SOLUTION

Each person spent a square sum of money, and each spent 63 g more than his wife. Hence M2W2 = 63. This means that M and W are integers satisfying

               (M + W)(M – W) = 63 x 1 or 21 x 3 or 9 x 7.

Solving for M, W shows that the number of pigs bought by the couples was (first two columns of the table below):

M1
8
Cornelius
W1
1
Gertrude
M2
32
Henry
W2
31
Anna
M3
12
Ely
W3
9
Catherine
We now use the information that Henry bought 23 more pigs than Catherine and Ely 11 more than Gertrude, to get column 3 of the table.

This gives Cornelius – Gertrude, Henry – Anna and Ely – Catherine.

Perhaps olde-time ladies didn’t have much to do with their spare time!


EXTENSION

This problem is based on the equation
X2Y2 = N, where X, Y, N are integers and where there are several ways of factoring N. Try constructing a problem of your own. For example, you might have a path bounded by two similarly placed squares with the same centre. The area of the path is 45 square metres, say (45 = 1 x 45 = 3 x 15 = 5 x 9). Now add some further constraints to get a single solution for X and Y.