I thought it would be more challenging if my two grandsons had a time-limit on their quest for squares and diamonds derived by subtraction (see above). Following a brief discussion, I lost and they negotiated a 9-minute time limit.
Why do I say “I lost” ? Well, I have a 7-minute “hour glass” a 4-minute “egg-timer” so, using these implements to measure 9-minutes, was going to be a bit of a Brain Teaser in itself
I wanted the boys to have as much subtraction practise as possible so, using these two timing pieces, how could I measure 9-minutes most efficiently ?
Mrs R had a go at this, but found that drawing 10 levels of squares and diamonds was quite tricky. She tried to draw it in the margin of her notepad, claimed she had a neat idea how to solve it, then left it to me. That gave me an idea for a spin-off teaser:
If the large square is 10 cm by 10 cm, what will be the size of the tenth shape?
Anyway, back to the original teaser. I skipped the square/diamond presentation and set out the four numbers as a row (in Excel). Set column E to be the same as column A to allow a full cycle. Each successive row has the formula A2 = abs(B1-A1) copied across columns A to D and down to row 10. Row 2 represents the second square etc. Then I used trial and error to find:
The reduction to four zeros will work with negative numbers as well., so I am told.
And with irrational numbers such as ‘Pi’
In fact, virtually any set of numbers … I am told there is only one set of numbers (all irrational) that will not reduce to zero. I’m afraid I had to cheat, and look that set of numbers up !
Hmm, this teaser is one of the toughest. Here’s my partial solution so far.
My thinking is that to go on indefinitely it must reach some stable state. I tried exponentials: in calculus, differentiating e^x leads to e^x which can go on indefinitely. No obvious joy there, unfortunately. I then went with the increasing differences from the original solution, 1,3,6,12, adding random small numbers, followed by manual tweaks until it started to settle down with more squares. I then scaled up to convenient integers to avoid the risk of rounding errors (quite a risk when repeatedly differencing).
Current best rational solution, achieving 39 squares:
78,347,480
206,125,821
275,597,501
313,368,490
It is starting to show some patterns, but I don’t recognise any of the numbers. I have a few ideas what to try next. It will have to wait as I am building a shed for my sister for the next couple of days.
This teaser is a derivative of one of the earliest on the Forum, probably more than 20 years ago!
My dear departed friend, the Grand Duke of Newbury, had a Treasure House. Each room in this Treasure House contained as many Treasure Chests as there were Treasure Rooms in the Treasure House. And each Treasure Chest contained as many diamonds as there were Treasure Chests in that Treasure Room.
Now, the Grand Duke had arranged with the Executor of his Will, that on his departure from this life, the Executor should provide his brother, the Grand Duke of Salisbury, with one Treasure Chest of diamonds and that the remaining diamonds be divided equally among his six sons. Any remaining diamonds were to be placed in a delicate ceramic casket and sealed. The sealed casket and its contents were bequeathed to my good-self.
I was relating this tale to my two young grandsons and showed them the said casket, which is still sealed and has never been opened. They were excited beyond belief and yelled out “How many diamonds are there in the casket ?”
“Well” said I, “how many diamonds do YOU think should be in the casket ? “
There is a 1 in 3 chance that she she chooses any one burger, same for the drink. The chance of any one combination is 1/3 x 1/3, or 1 in 9.
The chance that she does not choose any one burger is 2/3, and the same for the drink. The chance of no cheeseburger and no cola is 2/3 x 2/3, or 4 in 9, so there is a 5 in 9 chance that she orders at least one or the other.
Well, you know I like a challenge. I had another look at my 39 squares solution…
Adding or subtracting any constant from each of the four chosen numbers makes no difference to the final solution because the constant just cancels out. Hence subtract to make the first number equal to zero.
Similarly, scaling all the chosen numbers by a constant makes no difference other than scaling all the answers. Hence scale to make the second number equal to one. Now we only have two unknowns and estimates for them of 1.543689012 and 1.839286754. As expected, these generate 39 squares as before.
I then spotted a pattern that they were reducing steadily whilst shifting one step to the left. Taking ratios showed a near constant c = 0.8392868, uncannily related to the fourth number. Hence we can derive some formulae, assuming our chosen numbers are 0, 1, x, y:
c = y-1
From square #3:#2: 1.543689/1.839287 = x/y = c = y-1, hence x = y^2-y
From square #3:#2: 0.248091/0.295598 = (2x-y-1)/((y-x) = c = y-1
After a bit of algebra it gives us a fairly neat looking: y^3 - y^2 - y - 1 = 0
I had to Google how to solve cubic equations. After some tedious algebra it gives a fairly clunky looking answer:
y = ((19+3sqrt(33))^(1/3) + (19-3sqrt(33))^(1/3) + 1)/3 = 1.83928675521416113255…
But my initial solution was pretty close! Here is a summary of some of the workings…
I plugged it in to my enhanced precision spreadsheet and achieved 10,000 squares, though that did require over 2,600 decimal places to ensure it remained stable!
You got it ! Well done Ravvie. The source I found used ‘a’ so a^3 - a^2 - a - 1 = 0
(Of course, any letter will do. !)
To keep things simple, the 4 starter numbers that I was given were 1, a, a^2 and a^3 which to two decimal places are
1
1.84
3.38
6.22
But without a super enhanced precision spreadsheet I have had to trust others that these numbers result in a never ending set of squares, (even though they do converge, allegedly)
Some of the recent teasers are the result of helping my 16 year old grandson revise for his GCSE this summer. I have tried to ‘dress them up a bit’ to disguise the fact that they are basic maths. So apologies.
However, one of his practice revision papers (calcuators are NOT allowed) includes the following - “evaluate 8^(1/3)”
ie “eight to the power one-over-three”
He had problems with this, so I explained, using a more general example, that 8^(2/3) simply means "the square, of the cube root, of 8.
Hence 8^(1/3) ie the practise paper example, simply means “the cube root of 8” which of course is 2.
He is able to check his answers at school. Three times I’ve asked him to do so, and each time he tells me that the answer given by the school in this case is 0.5.
I have checked on my calculator and in my ancient school maths books and still consider the answer to be 2.
Is anybody confident enough to be positive about the correct evaluation ?
PS. this is not a “Brain Teaser” per-se, simply a genuine enquiry so as not to give my grandson wrong info, and/or to put my mind at rest !!
