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)

Posted on: 15 December 2018 by Don Atkinson
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. 

Posted on: 15 December 2018 by Ian G.
Don Atkinson posted:
Ian G. posted:

13325 units for A to F

It’s less than 13325 miles

Second iteration..... 5525 miles 

Posted on: 15 December 2018 by fatcat
Don Atkinson posted:
fatcat posted:

20

It’s more than 20.

I know. A classic case of not taking enough care reading the question.

Posted on: 15 December 2018 by Don Atkinson
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.

Posted on: 15 December 2018 by Don Atkinson
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 !

Posted on: 15 December 2018 by Ian G.

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... !)

Posted on: 15 December 2018 by Don Atkinson
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 !!

Posted on: 15 December 2018 by Don Atkinson

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 !

Posted on: 15 December 2018 by Filipe
Filipe posted:
Don Atkinson posted:

Parallelogram Angles Jpeg

If the double slashes mean equal length then 58

Phil

Didn’t get feedback on this, Don.

Posted on: 15 December 2018 by Don Atkinson
Filipe posted:
Filipe posted:
Don Atkinson posted:

Parallelogram Angles Jpeg

If the double slashes mean equal length then 58

Phil

Didn’t get feedback on this, Don.

Oh dear, I know that I responded, but that response is clearly missing.

the double slashes do mean equal length and the angle is 58 deg.

Well done !

Posted on: 15 December 2018 by Filipe
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

Posted on: 15 December 2018 by Filipe
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

Posted on: 16 December 2018 by Don Atkinson
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.

Posted on: 16 December 2018 by Filipe

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

Posted on: 16 December 2018 by Don Atkinson

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.

Posted on: 16 December 2018 by Don Atkinson

Oh! and the hypotinuse is more than 50.

Posted on: 16 December 2018 by Don Atkinson

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 !!

Posted on: 16 December 2018 by Don Atkinson

It's more than 5 !

Posted on: 16 December 2018 by Filipe

I corrected my answer to 5 on the full circle.

Posted on: 16 December 2018 by Don Atkinson
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 !

Posted on: 16 December 2018 by Filipe

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.

Posted on: 16 December 2018 by Don Atkinson
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 !

Posted on: 16 December 2018 by Filipe

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.

Posted on: 16 December 2018 by Filipe

Hope half miles are not allowed on the first leg!

Posted on: 16 December 2018 by Filipe
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