Brain Teasers ? or 50 Years On........... ?

Posted by: Don Atkinson on 02 June 2015

50 Years on…….

 

50 years ago, I was doing what many 18 year olds are doing this week and over the next few weeks……………….their A-Levels.

 

Mine were Pure Maths; Applied Maths; Physics and Chemistry. We also had a new subject called The Use of English.

 

About 10 years ago I started a few “Brain Teaser” threads on this forum. One or two people complained that many of the so-called Brain Teasers were no more than A-Level maths dressed up. That was true of a few teasers, but most were real teasers, especially the ones like “The Ladder” posted by Bam and also the one about the maximum number of 1cm diameter spheres that can be packed into a 10x5x5 cm box.

 

Any way, never mind Brains or Teasers, I guess one or two other Forumites are also looking back 50 years and would be delighted to tease their brains with calculus, probability, spherical geometry, geometric progressions, Newton’s Laws of Motion ……………………….no ? Then probably best if you drink your weekly 21 units tonight and wake up in the Music Room tomorrow to recover from the nightmare !

 

First one to follow shortly, and please, please add your own favourites !!

Posted on: 24 June 2015 by Don Atkinson
Originally Posted by Simon-in-Suffolk:

Ok my turn, and a variant of a topical question..

 

Derek has a bag of marbles, with three colours; blue, green and yellow marbles in it.

The bag contains 5 green marbles.

Derek finds the chance of him taking just two green marbles out of the bag is 10/21

How many marbles are in the bag?

 

Simon

Neat one Simon

Posted on: 24 June 2015 by Don Atkinson
Originally Posted by fatcat:
Originally Posted by sophiebear0_0:

Fatcat

 

I think your solution can be simplified to D^2 / L^2 (ie same as Don's)

 

So I am now out-numbered !

 

But I'm still holding glimmer of hope for my approach.

 

Regards,

 

Peter

I see what your saying Peter.

 

When you look at one square on my diagram, the ratio of shaded/unshaded corresponds to Dons formula.

 

Do we have a  thre-way consesus yet ?

 

Is it "Great minds....." or "Fools seldom....." ?

Posted on: 24 June 2015 by sophiebear0_0

Don

 

I'm pretty certain that both you and Fatcat are correct. Well done !

 

I'm still not 100% certain where my logic breaks down - which is a little frustrating. Just to try and clarify my original train of thought:

 

1) For the original problem where the disc was confined to a square, the disc could not pass any boundaries.

 

2) For the revised problem where the disc was confined to no more than 2 squares, it could only pass a single boundary.

 

3) If the disc passes a single boundary, the only possible configuration is that it resides within a rectangle of dimensions 2L * L

 

4) I then worked out the probability for a disc to land in a rectangle 2L * L

 

5) I then took 1 minus the probability calculated in (4) above.

 

Voila. All very elegant, it made sense to me.........but its clearly wrong !!!!

 

Well done gents, I think your solution is equally elegant. And it has the advantage of being correct !

 

Best regards,

 

Peter

 

And thanks Winky for getting us all to put our thinking caps on.

Posted on: 24 June 2015 by Don Atkinson
Originally Posted by sophiebear0_0:

Don

 

I'm pretty certain that both you and Fatcat are correct. Well done !

 

I'm still not 100% certain where my logic breaks down - which is a little frustrating. Just to try and clarify my original train of thought:

 

1) For the original problem where the disc was confined to a square, the disc could not pass any boundaries.

 

2) For the revised problem where the disc was confined to no more than 2 squares, it could only pass a single boundary.

 

3) If the disc passes a single boundary, the only possible configuration is that it resides within a rectangle of dimensions 2L * L

 

4) I then worked out the probability for a disc to land in a rectangle 2L * L

 

5) I then took 1 minus the probability calculated in (4) above.

 

Voila. All very elegant, it made sense to me.........but its clearly wrong !!!!

 

Well done gents, I think your solution is equally elegant. And it has the advantage of being correct !

 

Best regards,

 

Peter

 

And thanks Winky for getting us all to put our thinking caps on.

hmmmm,

 

I think I can see your line of reasoning. I'll have a careful look over the next day or so, but its awfully difficult to rationalise good ideas that somehow, just don't quite seem right, but who knows.............!

 

Cheers

Don

Posted on: 24 June 2015 by Don Atkinson
Originally Posted by sophiebear0_0:

Don

 

I'm pretty certain that both you and Fatcat are correct. Well done !

 

I'm still not 100% certain where my logic breaks down - which is a little frustrating. Just to try and clarify my original train of thought:

 

1) For the original problem where the disc was confined to a square, the disc could not pass any boundaries.

 

2) For the revised problem where the disc was confined to no more than 2 squares, it could only pass a single boundary.

 

3) If the disc passes a single boundary, the only possible configuration is that it resides within a rectangle of dimensions 2L * L

 

4) I then worked out the probability for a disc to land in a rectangle 2L * L

 

5) I then took 1 minus the probability calculated in (4) above.

 

Voila. All very elegant, it made sense to me.........but its clearly wrong !!!!

 

Well done gents, I think your solution is equally elegant. And it has the advantage of being correct !

 

Best regards,

 

Peter

 

And thanks Winky for getting us all to put our thinking caps on.

I have highlighted where I think the flaw lies.

 

For the disc to overlap only one or two boundaries it's centre must be confined to a rectangle 2L*(L-d), not 2L*L as quoted in point 3.

 

But also it's centre could lie outside the long edges of this smaller rectangle provided it's centre is also more than d/2 from a short edge.

 

When extrapolated over the chess board, your concept looks like mine.

 

Your arithmetic then looks like 1 - [2L(l-d) -2(L-d)d]/[2L^2]

Which reduces to d^2/L^2

 

I think !

Posted on: 25 June 2015 by sophiebear0_0

Don

 

Many thanks for taking the time to try and determine where my reasoning breaks down. I'll try and unravel it all in the next few days when my head clears from discs & squares !

 

I probably wasn't clear about Point (3). I had assumed that the disc sits in a rectangle 2L * L but did calculate the probability based on the disc centre being confined to a rectangle  (2L-d) * (L-d). This gave rise to my original expression (after simplification).

 

I did appreciate that the disc could lie outside the long-side of the rectangle - but then just reconciled that the disc must still be inside a rectangle 2L * L, albeit a different rectangle. But I wasn't really comfortable with this - and I'm pretty sure that is where my logic has broken down. So thanks for pointing this out.

 

Thanks again !

 

Peter 

Posted on: 25 June 2015 by fatcat

Peter.

 

I've been drawing a few sketches, (I'm seeing squares and discs in my sleep)

 

I'm thinking along the lines the disc must land on at least one square but there is a possibility it could land on any of the four adjacent squares and only cross one line.

 

I've crunched a few number based on your theory concerning only crossing one line on the area shown in sketch below, but not come up with an answer.

 

 

Posted on: 25 June 2015 by Don Atkinson

Two Squares form a “strip” area 2L*L

 

If the centre of the disc lands within the “central strip” of these two squares it will never overlap more than two squares. Size of this “central strip” is 2L*(L-d)

 

If the centre of the disc lands outside this central strip, it might overlap more than two squares. But only if the centre of the disc lands within a distance of d/2 from the corner of each square.

 

If the centre of the disc lands outside this central strip, and further away from a corner than d/2  it can only overlap  two squares. There are four such areas, each (L-d)*d/2.Note, (Fatcat's diagram above shows a disc occupying one such area in the picture on the right)

 

Hence the total area in which a disc can land and overlap 0,1 or 2 squares is [2L*(L-d) + 4[(L-d)*d/2]

 

The Probability of a disc landing within this area therefor is  [2L*(L-d) + 4[(L-d)*d/2]/[2L*L]

 

The probability of a disc NOT landing within this area is 1 -  [2L*(L-d) + 4[(L-d)*d/2]/[2L*L]

 

This last expression looks complicated but it easily reduces to d^2/L^2

 

I’ll try to post a drawing later.

Posted on: 25 June 2015 by alainbil
Originally Posted by Don Atkinson:

Two Squares form a “strip” area 2L*L

 

If the centre of the disc lands within the “central strip” of these two squares it will never overlap more than two squares. Size of this “central strip” is 2L*(L-d)

 

If the centre of the disc lands outside this central strip, it might overlap more than two squares. But only if the centre of the disc lands within a distance of d/2 from the corner of each square.

 

If the centre of the disc lands outside this central strip, and further away from a corner than d/2  it can only overlap  two squares. There are four such areas, each (L-d)*d/2.Note, (Fatcat's diagram above shows a disc occupying one such area in the picture on the right)

 

Hence the total area in which a disc can land and overlap 0,1 or 2 squares is [2L*(L-d) + 4[(L-d)*d/2]

 

The Probability of a disc landing within this area therefor is  [2L*(L-d) + 4[(L-d)*d/2]/[2L*L]

 

The probability of a disc NOT landing within this area is 1 -  [2L*(L-d) + 4[(L-d)*d/2]/[2L*L]

 

This last expression looks complicated but it easily reduces to d^2/L^2

 

I’ll try to post a drawing later.

There is a simple way to obtain your result.  

 

If d> L, the hoop always lay across more than two LxL squares.

 

Otherwise, consider the LxL square in which the hoop center lays: The hoop will cross more than two squares if it center lays inside one of the four dxd squares located at the corners of the big square.

 

Then the probability that the hoop lays across more than two LxL squares is

 4(d/2)^2/L^2=(d/L)^2

 

Posted on: 25 June 2015 by Don Atkinson
Originally Posted by alainbil:
Originally Posted by Don Atkinson:

Two Squares form a “strip” area 2L*L

 

If the centre of the disc lands within the “central strip” of these two squares it will never overlap more than two squares. Size of this “central strip” is 2L*(L-d)

 

If the centre of the disc lands outside this central strip, it might overlap more than two squares. But only if the centre of the disc lands within a distance of d/2 from the corner of each square.

 

If the centre of the disc lands outside this central strip, and further away from a corner than d/2  it can only overlap  two squares. There are four such areas, each (L-d)*d/2.Note, (Fatcat's diagram above shows a disc occupying one such area in the picture on the right)

 

Hence the total area in which a disc can land and overlap 0,1 or 2 squares is [2L*(L-d) + 4[(L-d)*d/2]

 

The Probability of a disc landing within this area therefor is  [2L*(L-d) + 4[(L-d)*d/2]/[2L*L]

 

The probability of a disc NOT landing within this area is 1 -  [2L*(L-d) + 4[(L-d)*d/2]/[2L*L]

 

This last expression looks complicated but it easily reduces to d^2/L^2

 

I’ll try to post a drawing later.

There is a simple way to obtain your result.  

 

If d> L, the hoop always lay across more than two LxL squares.

 

Otherwise, consider the LxL square in which the hoop center lays: The hoop will cross more than two squares if it center lays inside one of the four dxd squares located at the corners of the big square.

 

Then the probability that the hoop lays across more than two LxL squares is

 4(d/2)^2/L^2=(d/L)^2

 

Hi Alainbil,

 

Ah, yes. If you look further up this page and back to the previous page, you will see that d^2/L^2 was posted as a solution.

 

However, Sophiebear had tried a different approach and we have been trying to help him understand the link between his solution and the one you mentioned and which, as you will see, others have also mentioned.

 

However, many thanks for providing further evidence, which I hope will help Sophiebear reconcile the two approaches.

 

Cheers

Don

Posted on: 25 June 2015 by sophiebear0_0

Gents

 

Thanks for you help. I now feel fully reconciled !

 

Thanks particularly Don for clarifying the gaps in my initial approach. It had been bugging me, knowing that I was wrong, but not fully understanding why. Your analysis has shed the light.

 

I can now sleep comfortably without thinking about discs in squares (or rectangles).

 

Regards,

 

Peter

Posted on: 26 June 2015 by Don Atkinson
Originally Posted by sophiebear0_0:

Gents

 

Thanks for you help. I now feel fully reconciled !

 

Thanks particularly Don for clarifying the gaps in my initial approach. It had been bugging me, knowing that I was wrong, but not fully understanding why. Your analysis has shed the light.

 

I can now sleep comfortably without thinking about discs in squares (or rectangles).

 

Regards,

 

Peter

Forget the squares and rectangles Peter and stick to Winky's circles. You got that one sorted in double-quick time !

Posted on: 26 June 2015 by Don Atkinson

PINs, ever forgotten yours for a moment ?

 

Mrs D forgot her PIN the other day. All she can remember is that at least one of the digits is a zero. The digits can take the value of 0, 1, 2.........8, 9.

 

What is the maximum number of PINs she will have to try ?

Posted on: 27 June 2015 by sophiebear0_0

Don

 

I'm a glutton for punishment !

 

I think it might be 3997 ?

 

Regards,

 

Peter

Posted on: 27 June 2015 by Don Atkinson
Originally Posted by sophiebear0_0:

Don

 

I'm a glutton for punishment !

 

I think it might be 3997 ?

 

Regards,

 

Peter

Hi Peter,

 

Sort of the "Same Parish" rather than the "Same Ball-Park" if you know what I mean. ie not the same number that I worked out.

 

But I might be wrong !

 

Cheers

Don

Posted on: 27 June 2015 by sophiebear0_0

Don

 

I think I see my error.

 

How about 3439 ?

 

Regards,

 

Peter

Posted on: 27 June 2015 by Lionel

3439 is what I have:10 to power 4 - 9 to power 4

Posted on: 27 June 2015 by Simon-in-Suffolk

I have 3240

 

That is 4x 10x9x9

 

Simon

Posted on: 28 June 2015 by Don Atkinson

The correct answer is 3439 and Lionel has shown us one of the two ways that I have used to solve this problem. It is the easier of the two ways.

 

Basically 10^4 is the combination of ALL the numbers, including one, two, three or four zeros. 9^4 is the combination of all the numbers  EXCLUDING the use of zero. The difference is all the combinations that include at least one zero.

Posted on: 01 July 2015 by alainbil
Originally Posted by Don Atkinson:

The correct answer is 3439 and Lionel has shown us one of the two ways that I have used to solve this problem. It is the easier of the two ways.

 

Basically 10^4 is the combination of ALL the numbers, including one, two, three or four zeros. 9^4 is the combination of all the numbers  EXCLUDING the use of zero. The difference is all the combinations that include at least one zero.

Another way is:

 

Either the first digit is 0, this gives 10^3 pins.

 

Or the second digit is zero: this gives 9 * 10^2 new pins (the case where the first digit is zero has been already counted).

 

Or the third is zero: this gives 9^2*10 new pins. ((the cases where either the first or the second digit is zero have been already counted).

 

Or the last one is zero: this gives 9^3 new pins.

 

Both ways give the same result since 10^4-9^4= 10^3+9 * 10^2+9^2*10+9^3

 

 

Posted on: 01 July 2015 by Don Atkinson

and now alainbil has shown us the other way

 

well done guys.

Posted on: 01 July 2015 by Don Atkinson

Going back to the idea of an "elliptical" fence around the lorry park..............

 

the equation for an ellipse is (x/a)^2 + (y/b)^2 = 1     where a and b are the major and minor axes respectively

 

If a = b we get a circle and In all these cases the centre of circle or ellipse is at 0,0

 

  • What shape do you get if you put a=3 and b=2 AND instead of using ^2 (to the power of two) you use ^ 2.5 (to the power of two point five) for both the "x/a" and "y/b" elements.ie (x/3)^2.5 + (y/2)^2.5 = 1
  • What shape do you get if you put a=3 and b=2 AND instead of using ^2 (to the power of two) you use ^ ∞ (to the power of infinity) for both the "x/a" and "y/b" elements. ie (x/3)^∞ + (y/2)^∞ = 1
Posted on: 01 July 2015 by fatcat

Time for a hifi related brainteaser.

The image below shows the measured frequency of a 1Khz test tone on the HiFi Sound, stereo test record.
(Ignore the fact the results aren’t even close to 1Khz, that’s an other story).

The frequency of the peaks and troughs indicate the fluctuating speed is due to the spindle hole being off centre.

If we say the maximum frequency is 988khz, the minimum is 980Khz and the average is 984Khz, is it possible to calculate how far the hole is off centre.

Posted on: 04 July 2015 by sophiebear0_0

Don

 

I have not been able to find any method to solve your latest offering, other than to crunch the numbers and plot the results ? The results are therefore far more interesting than the method. I get the following:

 

a) For the "raised to 2.5 power" problem the result is a curve similar to an ellipse (albeit a little bloated). It exists only in the quadrant for positive X and positive Y values. It is "centred on the origin (0,0) and has a major axis of 3 and minor axis of 2 (ie intersects x-axis at (3,0) and y-axis at (0,2)). The curve sits between a true ellipse (major axis = 3, minor axis =2) and a rectangle of X=3: Y=2. The curve forms an asymptote a it approaches (0,2) and (3,0).

 

b) For the case as the power approaches infinity, the resulting curve is a rectangle X=3; Y=2 with the  bottom left hand corner at (0,0).  Again I think this can only exist in the domain of positive X and positive Y ?

 

Regards,

 

Peter

Posted on: 04 July 2015 by alainbil
Originally Posted by sophiebear0_0:

Don

 

I have not been able to find any method to solve your latest offering, other than to crunch the numbers and plot the results ? The results are therefore far more interesting than the method. I get the following:

 

a) For the "raised to 2.5 power" problem the result is a curve similar to an ellipse (albeit a little bloated). It exists only in the quadrant for positive X and positive Y values. It is "centred on the origin (0,0) and has a major axis of 3 and minor axis of 2 (ie intersects x-axis at (3,0) and y-axis at (0,2)). The curve sits between a true ellipse (major axis = 3, minor axis =2) and a rectangle of X=3: Y=2. The curve forms an asymptote a it approaches (0,2) and (3,0).

 

b) For the case as the power approaches infinity, the resulting curve is a rectangle X=3; Y=2 with the  bottom left hand corner at (0,0).  Again I think this can only exist in the domain of positive X and positive Y ?

 

Regards,

 

Peter

I guess that Don meant

 

(|x|/3)^2.5 + (|y|/2)^2.5 = 1

 

and

 

(|x|/3)^∞ + (|y|/2)^∞ = 1

 

where |...| is the absolute value