Henry Ernest Dudeney/Puzzles and Curious Problems/50 - Buying Ribbon/Solution

by : $50$

 * Buying Ribbon

Solution
Mrs. Jones.


 * Hilda Smith and her mother bought $4$ yards and $8$ yards, paying $4 \oldpence$ and $1 \shillings 4 \oldpence$ respectively.


 * Gladys Brown and her mother bought $6$ yards and $12$ yards, paying $9 \oldpence$ and $3 \shillings$ respectively.


 * Nora White and her mother bought $9$ yards and $18$ yards, paying $1 \shillings 8 \tfrac 1 4 \oldpence$ and $6 \shillings 9 \oldpence$ respectively.


 * Mary Jones and her mother bought $10$ yards and $20$ yards, paying $2 \shillings 1 \oldpence$ and $8 \shillings 4 \oldpence$ respectively.

Proof
From this line:
 * and each person bought as many yards of ribbon as the number of farthings she paid for each yard

we get that:
 * the amount paid is a square number of farthings
 * the amount paid is in fact the square of the number of yards bought.

The mothers buy twice as much as the daughters, and hence the mothers buy an even number of yards.

Hence the amount the mothers pay is an even square.

It also follows that they pay $4$ times as much as their daughters do.

This problem is about differences between squares.

Nora bought $3$ yards less than an even number.

Hence Nora paid an odd square.

Similarly, the square numbers that are the purchases between Hilda and Mrs. Smith is $1 \shillings$ which is $12 \times 4 = 48$ farthings.

The parity of both purchases must match as their difference is even.

But as Mrs. Smith's payment is an even square of farthings, then Hilda's must be also.

This narrows down our choice of possible pairs of squares considerably.

Split $48$ in two, you get $24$ which is in the middle of $23$ and $25$.

Take the Odd Number Theorem and you know that you are looking at $11^2 + 48 = 13^2$ and again, both odd.

We also have $48 = 9 + 11 + 13 + 15$ which gives us $4^2 + 48 = 8^2$. Bingo.

So we know that Hilda bought $4$ yards of ribbon and paid $16$ farthings or exactly $4$ pence.

Her mother bought twice as much, that is, $8$ yards of ribbon for $64$ farthings or exactly $1 \shillings 4 \oldpence$.

So Mrs. Smith is indeed Hilda's mother, and they seem to be off to make something modest and unassuming for Hilda's presentation.

The differences paid between Mrs. Jones and Mrs. White equates to $\paren {12 + 7} \times 4 = 76$ farthings.

We want two square numbers whose difference is $76$, which is not a large set.

We already know that both those squares are even.

Let us start with $76 = 2 \times 38 = 37 + 39 = 20^2 - 18^2$

which is probably going to fit the bill.

Trying out $76 = 4 \times 19 = 16 + 18 + 20 + 21$ does not give us an AP of odd numbers, so that does not lead to a solution.

Nothing else can fit.

Hence we know that Mrs. Jones swanned out of the shop with $20$ yards of ribbon, for which she paid $400$ farthings, that is $8 \shillings 4 \oldpence$.

Poor Mrs. White had to slink out of the shop, embarrassed that she had been able to secure only $18$ yards of her own choice, for which she paid a mere $6 \shillings 9 \oldpence$.

We know that:
 * Mrs. Jones's daughter bought $10$ yards and paid $2 \shillings 1 \oldpence$ for it.
 * Mrs. White's daughter bought $9$ yards and paid $1 \shillings 8 \tfrac 1 4 \oldpence$ for it.

Nora can't be Jones or Smith, as they are even.

Suppose Nora is Brown.

Then Mrs. Brown bought $6$ yards, and her daughter (Nora Brown) bought $3$ yards.

But that gives us no mother buying $12$ yards, for we already know Gladys bought $6$ yards.

It follows that Nora cannot be Brown.

It follows that Nora is White.

Then:
 * Mrs. Brown bought $12$ yards, paying $144$ farthings, that is $3 \shillings$
 * her daughter bought $6$ yards, paying $36$ farthings, that is $9 \oldpence$

Gladys, we know, bought $2$ yards more than Hilda, so bought $6$ yards.

So if Nora is White, Gladys is Brown, and so Mary must be Jones.