Firstly, before emojis became a thing, many people used the semi-colon to show a wink. I could have said 64 as it hasn’t got a semi-colon after it. It was meant to demonstrate a silly reason for being the odd one out.
Secondly, “Going Back to my Roots” was a big hit in the UK and around the world in 1981. It was performed by the Soul/disco group “Odyssey”. I meant it to be a cryptic way of displaying the reasoning behind my answer - ie that 518 is the only number that doesn’t have integer square roots.
I’ll just stick to giving the answer in future!
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Ooops !
Sorry Steve. I wasn’t being critical of you or your answer. However, with hindsight it certainly looks like that. Please accept my apologies.
Your answer was good and very nicely disguised so as not to be a ‘spoiler’ for others. It is my clumbsy response and lack of knowledge about disco groups that is out of order. Again, my apologies.
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My apologies too Don; I wasn’t taking your reply as criticism. My attempt to be cryptic went a little too far, and I was just trying to explain my thought process.
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I didn’t get the Odyssey reference either. Maybe I’m a bit square.
I did get the winking emoji reference though.
At least I’m not the only one hijacking Don’s teasers with cryptic extensions. Keep them coming!
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No cryptic music references this time 
It seems to me that VW has the least, so I called that “V”, then used the other information to express each car in terms of “V”, eg Mercedes = V+18.
I then ended up with an equation of 6V+70=100, giving V =5.
Therefore the quantities are as follows:-
VW=5
Mercedes=23
Honda=17
BMW= 16
Audi=19
Ford=20
I’m sure there’s a much less laborious method; over to you Ravvie!
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Not really anything less laborious from me. I noted immediately that there were 7 pieces of information and 6 unknowns, so it would be solvable using any convenient simultaneous equations, and would have one redundant piece of information. It’s always worth checking there are sufficient pieces of information to avoid wasting time on an unsolvable problem.
I started with H as it appears in 3 equations giving:
A + F + M + H = (36-H) + (37-H) + (40-H) + H = 113 - 2H
Next:
113 - 2H + B + V = 100 = 113 - 2H + 21, so H = 17
The rest drop out quite quickly.
We didn’t need to be told there were 100 cars:
B = 39 - M = 39 - 40 + H = H - 1
V = 24 - A = 24 - 36 + H = H - 12
Adding these to 113 - 2H gives 100
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Another method that works is as follows:-
If we ignore the Honda + Ford =37 equation, the other 5 equations taken together include 5 of the vehicles twice, but the Ford not at all.
If we sum these other 5 equations, the sum of the 5 cars twice = 160. therefore the sum of them once must be 80, meaning that there must be 20 Fords (ie 100-80). It is then easy to work out the other cars’ values.
Not sure if this is any less laborious than mine and Ravvie’s earlier solutions, but an acceptable alternative I hope.
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Actually I think your approach of adding five equations is the neatest of the solutions so far. I like the lateral thinking.
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Well done Steve and well done Ravvie. Wonderful, elegant solutions and certainly not laborious !
I’m posting a different method below, which might help those whose first language isn’t Algebra 
On the other hand, I think this method qualifies as ‘laborious’.
It’s based on simple Trial and Error using arithmetic.
First, I re-ordered the pairs such that, as far as possible, the second car of each pair was named as the first car of the next pair. (not essential, but it helps to keep things in order)
Mercedes + BMW = 39
BMW + VW = 21
VW + Audi = 24
Audi + Honda = 36
Honda + Ford = 37
Mercedes +Honda = 40
I then Guessed (1st Trial) that there might be 25xMercedes. This would leave 14xBMWs.
14xBMWs would mean 7xVWs and consequently 17xAudi, 19xHondas and 18x Fords.
As a check, the Mercedes and Hondas would number 25 + 19 = 44.
44 doesn’t Fit with the Quoted Mercedes + Honda total of 40. (This is the first Error !)
So, let’s make a new guess (2nd Trial). Let’s try 23 Mercedes. (Yes, I know !!)
Working down the above list, as per the 1st Trial, would leave 16xBMWs, 5xVWs, 19xAudi, 17xHondas and 20xFords.
As a check, The Mercedes and Hondas would now number 23 + 17 = 40.
40 fits perfectly with the Quoted Mercedes + Honda total of 40.
Problem solved. No need for any more Trials or Errors.
Of course, it could easily take a few more Trials and Errors if that first guess was well wide of the mark ! But, if Algebra isn’t your passion …
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Algebra certainly isn’t my passion. I just use whatever maths tools are available (even inventing some illegal ones much to Mrs R’s disdain). I quite like the experimental approach.
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I agree with what Ravvie said, it is " the neatest of the solutions …"
Chopsticks
Today I decided to provide some support for my growing peas but the sticks I had were too long, being 4 feet. For the first stick, I chopped off 1 foot, being a quarter of its length, but it was still a bit too long. I then chopped off 3 inches, being a quarter of the previous cut. Still not right, so the next cut was 3/4 inch. I repeated this process for what seemed like forever.
For those that prefer to work in new-fangled metric units, I also had a stick that was 1200mm long at the start and I followed a similar process.
What was the final length of the stick? (Either imperial or metric is fine).
Just another 10 sticks to do, so I will be busy for a while!
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You could save a lot of time (well, forever) and a lot of sawdust, if you just cut a third off each original stick.
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I seem to recall a geometric formula for a converging series :
Sum = a/(1 – r)
In this case :-
a = 300mm
r = ¼
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Yes, the formula is sometimes known as a perpetuity. It was known and used by the Greeks and Chinese over 2000 years ago.
As an example, in the 3rd century, Chinese mathematician Liu Hui invented an algorithm for improving the effectiveness of Archimedes method for estimating π (drawing a hexagon in a circle and repeatedly doubling the number of sides to make the shape closer and closer to a circle). Liu Hui realised that the impact of each doubling of sides reduced by a factor of almost exactly four. He postulated that if such a pattern continued then he could use the perpetuity formula to predict that adding a third of the last increase would give a much closer result. He checked it using Archimedes method and found it doubled the number of decimal places.
So yes, using a third rather than repeatedly quartering would save a whole heap of work!
Unfortunately, Liu Hui’s insight and all his life’s work was lost for many centuries. On this particular topic, it took until the 17th century for Snell/Huygens to come up with a similar method. Just a few years too late to be useful for Ludolph van Ceulen who spent his life calculating π to 35 places using the slower method.
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Nice write-up Ravvie. Every day is a school day, as they say !
Well, I thought I would try the ‘slow’ method appropriate for the chopsticks.
Even using my squared paper, pencil and manual short division, one chop at a time, starting with 75mm, it only took five iterative steps (each dividing by four), to get to 99.98mm (10 decimal places) and to see where the convergence was leading.
I think that i’ll leave others to shoot 35 places
Mrs R’s Birthday Cake
This is a continuation of the previous theme.
I baked a chocolatey chocolate* cake for Mrs R’s birthday this week with chocolate filling and chocolate icing. I had used a square baking tin of some unknown size.
There are currently three of us at home, so how did I cut the cake to ensure everyone had exactly the same amount of all the cake?
I had no measuring devices nor squared paper for that matter! Just a straight sharp knife.
*The type of cake, although essential for birthday purposes, has no bearing on the mathematics of this teaser.
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I have a solution which I will draw in Powerpoint.
It assumes that the chef can …
Draw straight lines with his knife or at least cut the cake equally straight whilst holding its shape in the baking tin.
Use the knife and his finger to accurately transfer a measurement from one part of the cake to another part of the cake.
My solution starts with him marking (or cutting) the two diagonals.
Following on …
The diagonals intersect at the mid-point of the square.
He can transfer this mid-point to mark the mid-point of any of the four sides of the square.
He can then draw a line from that ‘edge-of -the- square’ mid-point to each of the opposite corners of the square.
These lines will intersect the initial diagonals 1/3 of the width of the square.
This will enable the chef to cut the cake into three equal strips.
I used an 8x8 square, and a bit of y= mx + c generated two straight lines …
y = 2x
y = 8 - 3x
These lines intersect at x = 8/3
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