That’s pretty impressive Ravvie. Nice, simple numbers. Easy to remember etc. Well done.
However, … A quick check with Excel shows the answer is 9.0000000000000100000000000
ie there is an unwanted ‘1’ in the 14th decimal place. Argh !!
Given Ravvie’s contribution above, I think we should acknowledge your early solutions containing negative numbers and very small rounding approximations.
Well done
I’ll post the solution that i’m aware of later this evening or tomorrow.
How anybody, starting from scratch with pencil and paper could come up with a solution, beats me.
One useful technique with rational numbers is to use integers as far as possible. That avoids rogue roundings.
Hence use 20^3 - 17^3 = 9x7^3 exactly
My approach was to simplify the problem. Start with:
a^3/c^3 + b^3/d^3 = 9
Multiply by c^3:
a^3 + b^3 x c^3/d^3 = 9c^3
As the first and third items are integers, the second one must be an integer also. Hence we can use c = d to give us:
a^3 + b^3 = 9c^3. Hence my cryptic reference to Fermat (& Wiles) as replacing 9 with 1 has no positive integer solutions per Fermat’s last Theorem.
It’s one equation with three unknowns. I assume that in 1900 the mathematician used some clever maths that either I have forgotten or more likely don’t know. Hence I used Excel with some smallish(?) positive numbers and could only find the initial solution that Don gave. Allowing say b to be negative quickly showed a single solution that is exact.
I re-did the Excel calculation, using a more rational approach eg 20^3 - 17^3 rather than (20/7)^3 etc
This eliminated that unwanted ‘1’ in the 14th decimal place.
The three positive integers, that will generate 9 are
a) 415,280,564,497
b) 676,702,467,503
c) 348,671,682,660
(c) is the denominator, so the two numbers are a/c and b/c hence
(a/c)^3 + (b/c)^3 = 9
Alternatively a^3 + b^3 = 9(c^3)
Remember, this was figured out using pencil and paper, well before computers came along. Astonishing IMHO 
And that, as I indicated initially, was the reason for posting this ‘teaser’
Hi Steve, Ravvie and (even more so) anyone else who might be browsing and interested …
… If you type into Google etc any one of those 12-digit numbers a), b) or c) that I quoted above, you should be able to get a few websites that provide a bit more about the history of the ‘teaser’
The ‘Dude’ who provided a solution in c.1907 was called … ?
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How old !!
Mrs D is convinced she is getting younger. (Of course she is, she occasionally reads these teasers).
This subject came about because our eldest grandson has just had a birthday.
Today, as ‘proof’ that she is getting younger, she pointed out that now, she is only four times as old as our eldest grandson is now, whereas five years ago she was five times as old as he was then.
I don’t have the courage to explain the fallacy in her argument ! But by now I’m guessing that some of you know what their current ages add up to ?
PS. Despite the answer to the teaser, she still only looks and acts as if she is a day or two older than twenty-one.
I think their current ages add up to the same number as a three dart maximum score minus MrsD’s current age.
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I agree with @SteveD’s cryptic answer, being a sum of 100 for their current ages. But you should tell Mrs D she looks 20.
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Well done Steve.
And a nice way of describing/concealing the answer so as to avoid spoiler.
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Well done Mike.
I did once try reversing the ages in the teaser, but she didn’t seem to be entirely convinced !
PS. For the avoidance of doubt, the ages etc are all ficticious.
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Apart from the age 21 reference?
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I found some background on Dudeney’s teaser. The original problem was a “doctor of physic” had two “spherical phials” of 1 and 2 foot circumference. He wanted to have the exact measures of two other phials, of a like shape but different in size, that may together contain just as much liquid as is contained by these two. Clearly 1907 speak.
I found an outline solution which I’m trying to fill in the details. His solution was to use geometry (and algebra, calculus and polynomial long division). No wonder it was tough! Visually (my diagram) it looks like this:
The blue curve is the cubic x^3 +y^3 = 9, where x and y are the two sizes. The key points are:
If we know two rational points (roots) of a straight line passing through the curve then the third point will also be rational.
A tangent counts as two roots, both the same.
The curve is symmetrical in that x and y are interchangeable (e.g. sizes 1 and 2 vs 2 and 1).
The aim is to find a rational point in the top right quadrant to give positive x and y. This is achieved by connecting any points known to be rational, then working out the third point.
The chart starts at point P and goes to Q, Q’, R and finally S. I can do up to R by hand but haven’t yet got to the final solution, S.
It is a tough one!
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Definitely looks more like advanced mathematics than a mere brain teaser! Impressive.
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Indeed, it is rather mind boggling.
It was published in a book called “Dudeney’s Canterbury Puzzles”, though I think this puzzle was considered rather fiendish at the time!
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I still don’t know what the answer is either!
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Hi Mike,
You can do the arithmetic ‘long hand’ to confirm it is ‘exact’, or …
… put the numbers into Excel, making sure that where appropriate, you have maximised the decimal places in Excel
I chose the second option, and it convinced me that the numbers are exact.
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Having just read wikipedia’s ‘Numeric precision in Microsoft Excel’, I note that 15 significant figures is about its limit. I’m no longer certain that Excel is adequate to perform the arithmetic using those 12-digit numbers !
Looks like we are now entirely dependent on ‘Blind Faith’ in Dudeney’s 1907 work or …
… Ravvie !
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