A Fistful of Brain Teasers
Posted by: Don Atkinson on 13 November 2017
A Fistful of Brain Teasers
For those who are either non-British, or under the age of 65………. The UK used to have a brilliant system of currency referred to as “Pounds, Shillings and Pence”. Simplified to £ ״ s ״ d. No! Don’t ask me why the “Pence” symbol is a “d”, just learn it and remember it !
A £ comprised 20 Shillings and a Shilling comprised 12 Pence. Thus a £ comprised 240 Pence. I reckon that both Microsoft and Apple would have difficulty with these numbers in their spreadsheets, more so if we included Guineas, Crowns, Half-Crowns and Florins. However, I digress..............
The purpose of the explanation is to assist with the first two or three teasers that follow. So just to ensure a reasonable comprehension has been grasped…. ….. if each of three children has £3 − 7s − 9d, then collectively they have £10 − 3s − 3d Got the idea ? Good ! Just try 5 children, two each with £4 − 15s − 8d and three each with £3 − 3s − 4d. How much do they have between them ? (this isn’t the first brain teaser, just the basic introduction with some “homework”, the Teasers follow)
winkyincanada posted:Don Atkinson posted:The two wheels on each axle of a railway wagon are rigidly connected together.
When an engine negotiates a bend one would expect one of the wheels to skid (because one wheel has to go
further than the other).
Neither wheel skids, however, even if the rails aren't banked.
Why?PS I don't think that the solution usually quoted completely eliminates wheel skid at all speeds, but it certainly plays a significant part.
The wheels are tapered like shallow truncated cones. They move relative to the track so the outside wheel has an effectively greater diameter than the inside wheel during a turn. Super elevation of the track (banking) assists a bit, but the taper is the main thing. The taper also helps stop the wheel flanges from hitting/rubbing the rails when the train is going straight.
Nice explanation winky.
And it’s right.
Don Atkinson posted:Ian G. posted:13325 units for A to F
It’s less than 13325 miles
Second iteration..... 5525 miles
Don Atkinson posted:fatcat posted:20
It’s more than 20.
I know. A classic case of not taking enough care reading the question.
Ian G. posted:Don Atkinson posted:Ian G. posted:13325 units for A to F
It’s less than 13325 miles
Second iteration..... 5525 miles
You are heading in the right direction Ian.
Just to be certain that I haven’t dropped a clanger, each distance AB, AC.........AF is an integer number of miles. Also, BF, CF etc are integer miles. AF is the smallest solution for AF.
fatcat posted:Don Atkinson posted:fatcat posted:20
It’s more than 20.
I know. A classic case of not taking enough care reading the question.
I know the feeling !
Don, I think your explanation is (and was) clear.
Final try before bed 1105 ( note to self - don't trust online lists of pythagorean triples and some are incomplete... !)
Ian G. posted:Don, I think your explanation is (and was) clear.
Final try before bed 1105 ( note to self - don't trust online lists of pythagorean triples and some are incomplete... !)
Well, you're on the right track and still heading in the right direction, but still someway to go !
And there are quite a few of those pythagorean triples around.
I think it's just a foot-slogging excercise unless there is one of those "Road to Damascus" moments. (there's no hidden meaning or hints in that last sentence) I only did it by trial and (lots of) error !!
I hadn't thought of looking up lists of triples. But I tried it just now and can see what you mean. Most lists don't contain all the triples.
However, I did find one list with the required triples and they were nicely listed together !
Don Atkinson posted:Ian G. posted:Don Atkinson posted:Ian G. posted:13325 units for A to F
It’s less than 13325 miles
Second iteration..... 5525 miles
You are heading in the right direction Ian.
Just to be certain that I haven’t dropped a clanger, each distance AB, AC.........AF is an integer number of miles. Also, BF, CF etc are integer miles. AF is the smallest solution for AF.
I read it as af, (ab + bf), (ac+cf), (ad+df), (ae + ef) all integer. Of course each may be integer also, and it may be the only way to solve it using Pythagorean pair.
Only just back from a party
Phil
Don Atkinson posted:Ian G. posted:Don, I think your explanation is (and was) clear.
Final try before bed 1105 ( note to self - don't trust online lists of pythagorean triples and some are incomplete... !)
Well, you're on the right track and still heading in the right direction, but still someway to go !
And there are quite a few of those pythagorean triples around.
I think it's just a foot-slogging excercise unless there is one of those "Road to Damascus" moments. (there's no hidden meaning or hints in that last sentence) I only did it by trial and (lots of) error !!
The Pythagorean triples satisfy the Euclid’s Formula (a, b, c) all integers
a**2 + b**2 = c**2
K**2(m**2 - n**2)**2 + K**2(2mn)**2 =K**2 (m**2 + n**2)**2
We are looking for lowest set of 4 (a,b) pairs with constant c
Need to do some programming. Certainly greater than 1000
Filipe posted:Don Atkinson posted:Ian G. posted:Don, I think your explanation is (and was) clear.
Final try before bed 1105 ( note to self - don't trust online lists of pythagorean triples and some are incomplete... !)
Well, you're on the right track and still heading in the right direction, but still someway to go !
And there are quite a few of those pythagorean triples around.
I think it's just a foot-slogging excercise unless there is one of those "Road to Damascus" moments. (there's no hidden meaning or hints in that last sentence) I only did it by trial and (lots of) error !!
The Pythagorean triples satisfy the Euclid’s Formula (a, b, c) all integers
a**2 + b**2 = c**2
K**2(m**2 - n**2)**2 + K**2(2mn)**2 =K**2 (m**2 + n**2)**2
We are looking for lowest set of 4 (a,b) pairs with constant c
Need to do some programming. Certainly greater than 1000
Hi Phil,
your pen-ultimate line is correct.
I can’t say the same about your final line.
Hi Don
Perhaps the page listing those below 1000 missed some! Bit lazy. I will have to get a list of prime numbers.
Unless of course we only look for the first pair and then reflect about the 90 degree axis! The answer being 50!
If you are really sneaky it’s 5
Phil
No reflections required.
It's just plain, straightforward, four lots of right-angled triangles with the same hypotinuse. And this hypotinuse and each of the other sides, (all eight of which are different), are all integers.
Oh! and the hypotinuse is more than 50.
I'm not sure if this process is called ITERATION or IRRITATION
20; 13325; 5525; 1105 ;1000; 50 ..........
but you are all closing the gap !!
It's more than 5 !
I corrected my answer to 5 on the full circle.
Filipe posted:I corrected my answer to 5 on the full circle.
I don't understand what you mean here !
Probably not your fault, just me, not understanding the obvious !
I’ve had my shower now! The two dissimilar triangles with hypotenuse 25 are the triples (7,24,25) and (15,20,25) swapping the first two adds (20,15,25) and (24,7,25).
The original question did not require that the sides are all different but did stipulate semicircle.
Filipe posted:I’ve had my shower now! The two dissimilar triangles with hypotenuse 25 are the triples (7,24,25) and (15,20,25) swapping the first two adds (20,15,25) and (24,7,25).
The original question did not require that the sides are all different but did stipulate semicircle.
True. However, the diagram shows four towns B, C, D and E. And these towns are clearly not two pairs reflected.
So, let's see.......how about a score of 5/10 for the situation to date ? and we'll keep 10/10 for the full monty of four different triangles (which will join the towns more or less as depicted in the initial diagram).
Well done so far !
The first four unique triples are (3,4,5), (5,12,13), (8,15,17), (7,24,25) so the product of their hypotenuses is 27,625.
Hope half miles are not allowed on the first leg!
Filipe posted:The first four unique triples are (3,4,5), (5,12,13), (8,15,17), (7,24,25) so the product of their hypotenuses is 27,625.
It’s 5,525 dividing by 5