# Strange mathematical riddle, from Dudeney's Amusements in Maths.

I was reading, and trying to solve, the following riddle by Dudeney's Amusements in Mathematics:

This puzzle closely resembles the last one, my remarks on the solution of which the reader may like to apply in another case. It was recently submitted to a Sydney evening newspaper that indulges in "intellect sharpeners," but was rejected with the remark that it is childish and that they only published problems capable of solution! Five ladies, accompanied by their daughters, bought cloth at the same shop. Each of the ten paid as many farthings per foot as she bought feet, and each mother spent 8s. 5¼d. more than her daughter. Mrs. Robinson spent 6s. more than Mrs. Evans, who spent about a quarter as much as Mrs. Jones. Mrs. Smith spent most of all. Mrs. Brown bought 21 yards more than Bessie—one of the girls. Annie bought 16 yards more than Mary and spent £3, 0s. 8d. more than Emily. The Christian name of the other girl was Ada. Now, what was her surname?

Now what one has to take into account is

1 y= 3 feet

1 farting = 1/4 of a dime (penny)

said this, and what I tried is this: writing the equation that rules the problem, by indicating with $x$ the price the Mrs pay, for $x$ feet, and with $y$ the price the daughter pay, per $y$ feet. Hence

$$x^2 = y^2 + 101.25$$

Where $101.25$ is the additional sum, IN PENNIES, that the mothers do pay.

Now, because the problem works in farthings, I actually converted as

$$x \frac{x}{4} = y\frac{y^2}{4} + 101.25$$

$$x^2 - y^2 = 405$$

With some "smart" algebra:

$$(x+y)(x-y) = 405$$

Hence by decomposing $405$ into possible products of two numbers $A$ and $B$ (with $A = x+y$ et cetera), I got those five couples of numbers:

$$\{1; 405\} ~~~ \{5; 81\} ~~~ \{3; 135\} ~~~ \{15; 27\} ~~~ \{9; 45\}$$

Now, for each of them, I solved the system

$$\begin{cases} x+y = A\\ x-y = B \end{cases}$$

Finding the two numbers which would give me the total $10$ price paid by mothers and daughters. For the records, those would be (respectilely for the couples of numbers found above):

$$x = 203, y = 202$$ $$x = 69, y = 66$$ $$x = 43, y = 38$$ $$x = 27, y = 18$$ $$x = 21, y = 6$$

When having to deal with the final part of the problem, the only that matches is

Mrs. Robinson spent 6s. more than Mrs. Evans

Where I can find she spent $27$ and Ms Evans spent $21$.

However, nothing else do match.

Where is my error?

• It would be helpful to express all those weird currency units in a common unit (and I'd suggest farthings) – Hagen von Eitzen Nov 26 '17 at 14:54
• @HagenvonEitzen Well yes that is what I did, for example all into farthings and all into feet. Yet it didn't help much... – Von Neumann Nov 26 '17 at 14:55
• I just researched the matter and it seems that 1 pound = 20 s, 1 s = 12 d, at least in the years 1066-1970? – Hagen von Eitzen Nov 26 '17 at 15:00
• @HagenvonEitzen Yes, that is correct! Indeed that book was written around 1909 ish – Von Neumann Nov 26 '17 at 15:08

You are doing fine so far. Each person's spending is the square of the numbers you quote. When Mrs. Robinson spent $6$s more than Mrs. Evans, that is $288$ farthings and $27^2-21^2=288.$ Now Mrs. Smith spent the most, so she bought $203$, Mrs. Jones spent about four times Mrs. Evans, getting about twice as much, so she bought $43$, leaving $69$ for Mrs. Brown. That means Bessie only bought $6$, so is Evans. Annie bought $48$ more than Mary, so Annie Brown bought $66$ and Mary Robinson $18$. Annie spent more than Emily, so Emily Jones bought $38$ leaving $202$ for Ada Smith.