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THE DUTCHMEN'S WIVES.

(Money Puzzles)
I wonder how many of my readers are acquainted with the puzzle of the
"Dutchmen's Wives"--in which you have to determine the names of three
men's wives, or, rather, which wife belongs to each husband. Some thirty
years ago it was "going the rounds," as something quite new, but I
recently discovered it in the _Ladies' Diary_ for 1739-40, so it was
clearly familiar to the fair sex over one hundred and seventy years ago.
How many of our mothers, wives, sisters, daughters, and aunts could
solve the puzzle to-day? A far greater proportion than then, let us
hope.
Three Dutchmen, named Hendrick, Elas, and Cornelius, and their wives,
Gurtruen, Katruen, and Anna, purchase hogs. Each buys as many as he (or
she) gives shillings for one. Each husband pays altogether three guineas
more than his wife. Hendrick buys twenty-three more hogs than Katruen,
and Elas eleven more than Gurtruen. Now, what was the name of each man's
wife?


Answer:

The money paid in every case was a square number of shillings, because
they bought 1 at 1s., 2 at 2s., 3 at 3s., and so on. But every husband
pays altogether 63s. more than his wife, so we have to find in how many
ways 63 may be the difference between two square numbers. These are the
three only possible ways: the square of 8 less the square of 1, the
square of 12 less the square of 9, and the square of 32 less the square
of 31. Here 1, 9, and 31 represent the number of pigs bought and the
number of shillings per pig paid by each woman, and 8, 12, and 32 the
same in the case of their respective husbands. From the further
information given as to their purchases, we can now pair them off as
follows: Cornelius and Gurtruen bought 8 and 1; Elas and Katruen bought 12
and 9; Hendrick and Anna bought 32 and 31. And these pairs represent
correctly the three married couples.
The reader may here desire to know how we may determine the maximum
number of ways in which a number may be expressed as the difference
between two squares, and how we are to find the actual squares. Any
integer except 1, 4, and twice any odd number, may be expressed as the
difference of two integral squares in as many ways as it can be split up
into pairs of factors, counting 1 as a factor. Suppose the number to be
5,940. The factors are 2 squared.3 cubed.5.11. Here the exponents are 2, 3, 1, 1.
Always deduct 1 from the exponents of 2 and add 1 to all the other
exponents; then we get 1, 4, 2, 2, and half the product of these four
numbers will be the required number of ways in which 5,940 may be the
difference of two squares--that is, 8. To find these eight squares, as
it is an _even_ number, we first divide by 4 and get 1485, the eight
pairs of factors of which are 1 x 1485, 3 x 495, 5 x 297, 9 x 165, 11 x
135, 15 x 99, 27 x 55, and 33 x 45. The sum and difference of any one of
these pairs will give the required numbers. Thus, the square of 1,486
less the square of 1,484 is 5,940, the square of 498 less the square of
492 is the same, and so on. In the case of 63 above, the number is
_odd_; so we factorize at once, 1 x 63, 3 x 21, 7 x 9. Then we find that
_half_ the sum and difference will give us the numbers 32 and 31, 12 and
9, and 8 and 1, as shown in the solution to the puzzle.
The reverse problem, to find the factors of a number when you have
expressed it as the difference of two squares, is obvious. For example,
the sum and difference of any pair of numbers in the last sentence will
give us the factors of 63. Every prime number (except 1 and 2) may be
expressed as the difference of two squares in one way, and in one way
only. If a number can be expressed as the difference of two squares in
more than one way, it is composite; and having so expressed it, we may
at once obtain the factors, as we have seen. Fermat showed in a letter
to Mersenne or Frenicle, in 1643, how we may discover whether a number
may be expressed as the difference of two squares in more than one way,
or proved to be a prime. But the method, when dealing with large
numbers, is necessarily tedious, though in practice it may be
considerably shortened. In many cases it is the shortest method known
for factorizing large numbers, and I have always held the opinion that
Fermat used it in performing a certain feat in factorizing that is
historical and wrapped in mystery.










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