Brain Teaser No 1

that they are in sequence is somewhat profound I guess

The hurt is receeding !!
fast !

Cheers

Don

PS it was only meant to be a 'stop gap' or 'stocking filler' whilst we filled out the details in your more interesting riddles puzzle.

2 forum members have decided to visit Don in Newbury and are heading north. Unfortunately they have got lost and confusion settles in. They approach a roundabout, 4 exits, the usual but there is some confusion so they end up going round the roundabout, having started at the same time. Funnily enough they have the same car. Driving as fast as they can ignoring any speed limits, one is in the inside lane (20m from the center of the roundabout) and the second is 1m further out, they exit the roundabout on the first exit after they finally meet and are lucky enough to be heading in the right direction.

Where was was the round about relative to Newbury.

Matthew

No, not a Naim recap, a Mathew recap.....

2 forum members have decided to visit Don in Newbury and are heading north....... They approach a roundabout.........Where was the round about relative to Newbury.

Cheers

Don

quote:

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Paul Ranson: The riddle problem has much unnecessary (dis)information.

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Well it seems Matthew has done it again with the last puzzle.

-Dan

I think you need all the information for this one.

OTOH given the impossibility of entering the roundabout side by side...

Anyway I guess the chap on the inside goes around 40.49 times, so they leave (rather sharpish) by the North exit which makes the roundabout South of Newbury.

I'll probably have changed my mind tomorrow.

Paul

Paul,

But they where heading north to start with....

Momentary panic when I thought maybe there wasn't a unique solution but a quick calculation found my fears to be unfounded.

Matthew

Matthew

I think the roundabout is north of Newbury.

The inside lap is 40*pi
The outside lap is 42*pi

When the inside car has done 1 lap, the outside car has done 40/42 of a lap.

When the inside car has done "x" quarter laps, the outside car has done (40/42)*x quarter laps.

The first time the cars meet at the same quarter, the difference between the number of quarters each has travelled must be 4.

Therefore x - (40/42)*x = 4

Therefore x = 84.

The inside car has done 84 quarters or 21 laps.
The outside car has done 80 qurters or 20 laps.

They therefore leave at the same exit at which they entered, and will be travelling south. The roundabout must therefore be north of Newbury.

A bit long-winded; presumably there is a neater answer.

Steve D

The cars aren't travelling at the same speed.

If you're travelling North then you enter the roundabout from the South, if you go around just under 40.5 (or 41.5) times then you must leave travelling North, heading for Newbury. I think.

Paul

Paul,

You are right. As I made up the brian teaser on the spot and hadn't actually found a solution (this is very bad form) I was trying to figure it out myself.

Don,

Happened to pass by Newbury yesterday on my way to Milton Hall thus the inspiration!

cheers

Matthew

Matthew & Paul,

The wording of your question suggests that the cars ARE going at the same speed.

I must admit that I had misread the puzzle initially, and assumed that they met AT an exit (ie a quarter-lap point). My solution was therefore more complicated than it needed to be.

A simpler solution is as follows (but gives the same answer).

If the inside car does "X" laps, the outside car does 20X/21 laps.

The inside car meets the outside car having done 1 more lap.ie

x - 20X/21 = 1

Therefore X = 21

Otherwise, same solution as before.

STEVE D

After, at first, thinking this was a trick question, I arrived at the same answer as Steve. If I understand the puzzle (which I appear not to do), the inner car needs to go one extra lap further than the outer car, i.e.
2pi(20)(x+1) = 2pi(21)x
which has the solution x = 20. They exit the same entrance as they entered. Now, if you mean to suggest the maximum velocity of each car is determined by the radius of curvature, well that's a different story.
cheers,
Dan

They're driving as fast as they can, and they have the 'same car' (it would be a tricky question if they were in the same car and the metre difference in radius was the distance between the front seats....)

Anyway I think that means they pull the same g, so,

(1) v1^2/20 = v2^2/21.

They're travelling for the same time, and the guy on the inside travels one more lap,

(2) (n+1)20v1 = n21v2

From (1) we have v1 in terms of v2,

(3) (n+1)20v1 = n21v1√(20/21)

(4) 20(n+1) = n√(20x21)

(5) n = 20/(√(20x21) - 20) = 40.49

Or maybe not.

Paul

quote:

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Originally posted by Paul Ranson:
(it would be a tricky question if they were in the same car and the metre difference in radius was the distance between the front seats....)

(5) n = 20/(√(20x21) - 20) = 40.49

Or maybe not.

Paul

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I should have thought of that, for this to work there would have to be some very thin vechicles. That was the solution I got to as well!

Matthew

quote:

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That was the solution I got to as well!

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Phew.

One day a beautiful problem will emerge fully formed in my mind. I hope.

Paul

After the success of the great amp giveaway, Naim USA decided to move some old pre-amps. This time, sealed crates contain a mixture of the following: 32's, 42's, and (as an incentive) 112's. The price of each amp is its product number in US$ (again dream on), so 32's cost $32, etc. Only the total price of each crate is known. Since, I really want a pre-amp with a remote, I'm hoping the crate labled $238- has a 112. Can I be sure?

cheers,
Dan
(hope this one is error free - first time cooking one up)

Dan, the crate labled $238 must contain 1x112 and 3x42. So take it with confidence.

Without a 112, the $238 price-tag would have to be a combination of 32s and 42s.

A simple spread-sheet does the arithmetic to show there are no combinations of (Ax32 plus Bx42) that can make 238. A and B are independant integers. Max value of A = 7, Max value of B = 5. A and B could be equal. A and/or B could be 0.

Cheers

Don

(hope this one is error free - first time cooking one up) So do I...........

Don,
That was fast! Yes, you have hit upon the solution - can you also show it is the only solution?
cheers,
Dan

Don,

Happened to pass by Newbury yesterday on my way to Milton Hall thus the inspiration!

cheers

Matthew

Local radio station mentioned two similar cars causing meyhem in Newbury at a busy roundabout. Also mentioned that police would like to hear from anybody who witnessed the event or who had inside information that might help them with their enquiries. The reward, IIRC, was £10,000. Looks like that NAP500 could soon be heading from Salisbury towards Newbury..........

Cheers

Don

Dan,

can you also show it is the only solution?

Fingers crossed, here goes......

There are ONLY 28 possible combinations of 32/42 that give results not more than £238.

The attached sheet highlights these 28 combinations.

It looks like a spreadsheet 'cos thats what it is. No gimicks, just easier and neater than doing the sums in my head and writing out the answers.

Of the WHOLE POPULATION of POSSIBLE combinations, ALL those to the east and south of the shaded area are bigger than $238 and hence eliminated. Note, I haven't shown the WHOLE population, I stopped at 9 in each direction.

The 28 remaining 'possibilities' can also be eliminated by conclusive 'inspection' ie none of them equals $238.

Now, I did think about showing there are no combinations of prime factors derived from 32 and 42 that generate 238, but I decided that simply inspecting 28 numbers was just as valid and a lot easier.

I doubt if anyone else will agree..........

Fingers uncrossed,

Cheers

Don

Don,

Didn't go into Newbury, honest!

Matthew

The roundabout problem hasn't been satisfactorily concluded.

Paul and Mathew are agreed on 40.49 revolutions prior to exit, whilst Steved and Dan are agreed on 20 for the outside car and 21 for the inside car.

I am with Steved and Dan on this one.

No fancy formulae. Mathew says the cars are the same and implies they are travelling at the same speed. The inside car travels 2(Pi)20 m in one lap and the outside car travels the same distance but is short of completing a full lap [by 2(Pi)m].

The outside car travels 2(Pi)21 m in one full lap.

After 21 laps, the inside car will have travelled 2(Pi)20x21 m. The outside car will have travelled the same distance. Since the outside radius is 21m, the figures are very conveniently arranged to show that the outside car has completed exactly 20 laps.

Tedious arithmetic, or a little spread sheet show this coincidence doesn't occour any sooner than 20/21 laps and that when it does occour it is EXACTLY on 20/21 laps so the cars leave by the same exit that they entered ie they will have reversed their direction.

Since they were travelling north when they entered the roundabout, they will leave heading south. Since this is the right direction for Newbury (according to the story line) the roundabout must be north of Newbury.

All those in favour say 'aye'

Cheers

Don

Tedious arithmetic, or a little spread sheet show this coincidence doesn't occour any sooner than 20/21 laps and that when it does occour it is EXACTLY on 20/21 laps so the cars leave by the same exit that they entered ie they will have reversed their direction.

Since they were travelling north when they entered the roundabout, they will leave heading south. Since this is the right direction for Newbury (according to the story line) the roundabout must be north of Newbury.

All those in favour say 'aye'

I re-read Mathew's original text and he says... the cars leave the roundabout at the first exit AFTER they first meet. Given that they first meet at the SAME exit at which they entered, then they would continue CLOCKWISE (well it is in the UK) for one quarter turn so heading WEST.

This would put the roundabout EAST of Newbury.

All those in favour shout 'aye'

Cheers

Don

My first answer was East of Newbury following your reasoning.

But if you read the question it says "Driving as fast as they can", not "the car on the outside drives at the same speed as the car on the inside".

So you have to make an allowance for this. The next level of assumption is that they can pull the same acceleration and determine the ratio between their speeds accordingly. It makes quite a difference to the outcome.

Now, we're making a big assumption that these cars handle well and are equally well setup...

Paul

I think we can all agree that the original question might have been more precisely worded. In fact Mathew suggested this a few posts back.

Perhaps Mathew would like to re-word the question?

The puzzle senario, as set, is impracticle to perform. eg the 1m difference in radii. However, two cars the same, each travelling at top speed.....implies they are travelling at the same speed IMHO, for the purposes of the puzzle.

Cars following a cicular path on a flat surface will tend to slide outwards. For a given speed this tendency will be greater at smaller radii. Friction between the road and tyres will prevent the slide, up to a limit. IMHO, the puzzle implies this limit isn't reached for either car. So the consequential implication is that each car is able to sustain its maximum speed, which IMHO as noted above, is the same.

If Mathew clarifies the question such that the outer car is 'just' on the point of sliding, then by implication the inner car would need to travel more slowley to avoid a slide. If, on the other hand, the inner car is just on the point of sliding, the outer car could easily maintain the same speed

Over to you Mathew...........

Cheers

Don

Don,

Again, nicely done with the solution. Since I didn't use a spread sheet you might be interested in my way of showing the solution is unique:

1) no combinations of 112's and 32's yield 238 since their greatest common divisor (16) does not divide 238 evenly.

2) the general solution of x*32 + y*42 = 238 is

x = 14 + n*42/2, y = -5 - n*32/2

which obviously has no solution with both x and y greater or equal to zero.

2) lastly, the general solution of x*112 + y*42 = 238 is:

x = 1 + n*42/14, y = 3 - n*112/14

and in this case only n=0 has a meaningful answer.

3) By brute force I found no combination of all amps totals $238-.

cheers,

Dan