A solution of the easy case is :
Well done alain, and nicely explained
Whilst the scale dilemma is being worked on another problem.
Apologies if this has already come up in the thread but have not been able to keep up with all the problems…
Here is a puzzle known as the Covent Garden Problem, which appeared in London, accompanied by the somewhat surprising assertion that it had mystified the best mathematicians of England:
Mrs. Smith and Mrs. Jones had equal number of apples but Mrs. Jones had larger fruits and was selling hers at the rate of two for a penny, while Mrs. Smith sold three of hers for a penny.
Mrs. Smith was for some reason called away and asked Mrs. Jones to dispose of her stock. Upon accepting the responsibility of disposing her friend’s stock, Mrs. Jones mixed them together and sold them of at the rate of five apples for two pence.
When Mrs. Smith returned the next day the apples had all been disposed of, but when they came to divide the proceeds they found that they were just seven pence short, and it is this shortage in the apple or financial market which has disturbed the mathematical equilibrium for such a long period.
Supposing that they divided the money equally, each taking one-half, the problem is to tell just how much money Mrs. Jones lost by the unfortunate partnership?
If X equals the number of apples each lady had originally, I deduced that Mrs Jones could have earned X/2 pence, and Mrs Smith X/3 pence, making 5X/6 in total.
Selling together, the total earnings were 4X/5, which was 7 pence less than 5X/6.
Solving the equation 5X/6 - 4X/5 = 7, gives X = 210.
Anyway, Mrs Jones was obviously a very poor businesswoman!
She ended up with her share of 84 pence compared to the 105 pence she could have earned on her own, for doing all the work - a 21 pence shortfall.
Mrs Smith on the other hand earned 84 pence for doing nothing at all, compared to 70 pence she would have earned on her own - a 14 pence surplus.
SteveD, a more elegant route to the solution than the one I came up with. Indeed Mrs Jones was a poor business women, or maybe overly generous!
Rather than SteveD’s elegant approach (well done Steve !) I hit it head-on with a rather crude bit of arithmetic.
I ‘assumed’ that each lady had 30 apples for sale. I chose 30 because it has prime factors of 2, 3, 5. These lined up with ‘two a penny’ ; ‘three a penny’ ; and ‘five a penny’.
Selling 2 x 30 (= 60) apples at the prices quoted, revealed a 1p shortfall when sold at ‘five a penny’.
For 7p shortfall, 30 x 7 = 210 apples for each lady would be needed. ie 420 apples in total.
The rest is more or less the same as Steve’s solution.
Hope this helps anyone whose algebra isn’t their strong point. 
Nice one Mathew ! (was your solution different to mine ?)
I have covered over the more difficult part - on the left hand side.
The right hand side illustrates the solution if the first weighing balances.
It is the same principle as described by alain, last week.
(you probably need to click the bottom of the image to make it large enough to read easily !)
I’ll start to reveal the left hand side this weekend.
The first couple of steps are crucial.
You can already see the first step, in which the two sets of balls do not balance, but you need to be certain to glean as much info as possible from this simple fact.
The second step is the bit that really is crucial to the whole enterprise !!
The third step(s) are relatively straight forward.
You can now see the 1st weighing in full.
If the scales balance, move to the right hand side and follow the process down to P, Q or R as appropriate.
If the scales don’t balance, follow the process down the left hand side. At the moment I have only revealed the first step in the left hand side.
The next step is important.
PS. Click on the bottom right corner of the diagram to enlarge it. Difficult to read otherwise !
I’ve revealed the crucial second step, assuming that on the first weighing, the scales didn’t balance.
I haven’t revealed the conclusions that I derived from this step.
It is easy now:
Key Safe Wordle Number
Last week, Mrs D insisted that I fit one of those Key Safes on the wall near our front door. I didn’t think it was a good idea, but when Mrs D insists …….
She dialled in a 4-digit PIN code that she was certain she could remember ………
OK:
The code is 4 digits, each appearing only once and, selected from the range 0 to 9 incl.
I have made four attempts at the code, as set out in the horizontal rows below.
The coloured circles to the right indicate how many digits are correct in that row.
A green circle indicates a correct digit in the correct location (Wordle anybody ?)
A yellow circle indicates a correct digit but in the wrong location
What was Mrs D’s Code ?
I didn’t use a thorough process tbh, but a bit of intuitive trial and error gave me 2395 as the code.
I used the same process … and got the same answer.
So we are either both wrong, or we are both right !!
