Countdown ?
What comes next in the following sequence ?
77 → 49 → 36 → 18 → ?
Countdown ?
What comes next in the following sequence ?
77 → 49 → 36 → 18 → ?
I have a free add-in to Excel, called XNumbers. It gives extended precision way beyond Excel’s 15 digit limit. It proves the answer is valid:
Clearly, doing this arithmetic by pen and paper would be rather tortuous!
I have just got back from the Etihad after celebrating a 6-1 win against Bournemouth so this is as much as I can cope with tonight. Maybe Blind Faith is in a bit more of a sober state!
Assuming I get Dudeney’s solution to work tomorrow, it should show that although his maths was rather hard, the arithmetic was somewhat easier, or at least a bit less daunting.
Wonderful. Well done Ravvie.
The limitations of Excel can easily be demonstrated by asking it to do
1 + A - 1
Substitute ‘2’ for ‘A’ and it gives you ‘2’
Substitute ‘0.0000000000001’ for ‘A’ and it gives ‘0’0000000000000999200…’
No doubt partly due to conversion to/from binary calculations.
What’s this new concept called ‘Artificial Intelligence’
More my speed, this one (assuming I’ve got it right)! I’ll say the next number is 8.
Bobby
On the limitations of Excel, I remember having lectures at University in which the professor encouraged us to write our own programmes in Fortran to avoid the limitations and propagation of errors that exist when using excel. This was 22 years ago, but it sounds like it’s still an issue.
Spot-on Bobby.
And it’s more my speed as well !
Dudeney’s Final Step
I finally got the method to work. Outline approach for the final phase of going from point R to the solution S (actually I flipped x and y):
Find the equation of the straight line of the tangent at R (needs differentiation etc).
Substitute this into x^3 + y^3 = 9 to give a new cubic (eliminates y).
Create a quadratic from the x values of the two known roots, both at R.
Divide the quadratic into the cubic to give the missing root (thanks to Mrs R for giving me a free 5 minute tutorial on polynomial long division).
Find the corresponding y value from the straight line equation.
Here’s my solution for the last two steps:
Whilst it is still rather complicated, at no time does Dudeney need to work with 36 digit figures. There are two divisions of 18 digit by 6 digit figures. Remember of course that this is to both generate and validate the unknown solution whereas previously we have just been validating a known solution.
Wonderful.
Well done Ravvie !
(and Mrs R for her polynomial tuition)
Milk Jugs …
Mrs D was doing some cooking this afternoon. She had an 8 pint jug that was full of milk and two empty jugs, one of 5 pint capacity and the other of 3 pint capacity. (ie three jugs in total)
She had two recipes. Each recipe required 4 pints of milk. (ie a total of 8 pints altogether)
Using only those three jugs, and being careful not to spill any milk, much to my surprise she poured the milk from jug to jug, several times and, hey-presto ! two lots of exactly four pints.
She had 4 pints in the 8-pint jug and the other 4 pints in the 5-pint jug.
Could you do this ?
And, just as importantly, can you describe your process clearly ?
PS. When I asked if she could do it again, she said “of course”. And did it a different way !!
-1.
Well done JR. I hope you enjoyed it.
And very good of you to avoid spoilers for others, who might also enjoy it.
Has anybody sorted out the milk jugs yet ?
After finally realizing that there are four small jugs (two respectively), I see two ways:
Assuming identical transparent jugs you completely fill one of the 5 pint jugs. The remaining 3 pints go into the other 5 pint jug. Then you take the one with more milk in it and pour some into the other one, until both contain the same amount. This should take some time, but at some point you will have divided the 8 pints nearly in half = 2 * 4 pints.
You completely fill one of the 5 pint jugs. Then you take that one and fill one of the 3 pint jugs, leaving 2 pints in the 5 pint jug. Then you completely fill the other 5 pint jug from the 3 and 8 pint jugs. Then you take the second 5 pint jug and fill the other 3 pint jug, leaving 2 pints in the second 5 pint jug, too. Now you just have to empty one of the 5 pint jugs, holding 2 pints each, into the other to get 4 pints. The remaining 4 pints of milk can now be poured into either the other 5 pint or the 8 pint jug.
Hi Mulberry,
That is a great response. Well done. However …
I thought the Teaser was clear, but I now realise that it needs to be made more clear - this is my mistake, not yours !!
There are only three jugs available. 8 pints, 5 pints and 3 pints
The jugs are not transparent and they are not necessarily similar in shape
No milk is lost.
You complete the task with 4 pints in the 8 pint jug and 4 pints in the 5 pint jug. The 3 pint jug is empty.
I apologise that the initial teaser is ambiguous. I’ll see if I can rectify that situation.
But, as I said above - well done.
I’ve changed the wording a little bit. Hopefully it’s now clear that there are only three jugs, and they are not necessarily of similar shape and they are not transparent.
I’ll post one of my solutions this evening to the 3-Jugs teaser. (I have two solutions).
As a hint, the first solution starts by filling the 3-pint jug, leaving 5 pints in the 8-pint jug.
Then pouring those 3 pints from the 3-pint jug into the 5-pint jug …
I hope I’ve read the instructions correctly.
errr … not quite ! But you have made a jolly good start with steps 1 to 3 inclusive…
Let me copy a couple of relevant text from (a) my initial post, which I slightly amended earlier today in response to Mulberry’s post and (b) from my reply to Mulberry.
(a)
(b) There are only three jugs available. 8 pints, 5 pints and 3 pints
The jugs are not transparent and they are not necessarily similar in shape
Oops…I read it a couple of weeks ago and misremembered the number of jugs!!