A Fistful of Brain Teasers

Not yet.

Perhaps others can ?

Division - aahhggg !!

Divide 45 into four parts such that the first part with two added, the second part with two subtracted, the third part multiplied by two and the fourth part divided by two, shall all be equal.

Thisshould  be a lot easier than arranging balls in boxes !

Don Atkinson posted:

Another Packing Problem

I have a box with internal dimensions 5cm x 5cm x 10cm and spherical balls of diameter 1cm. What is the maximum number of balls that I can fit into the box and close the lid?

Usual maths rules apply eg The balls are solid and won't squash. The box is solid and won't distort. All the dimensions are exact. ie 5 balls will just fit across the 5cm dimension, or 10 balls will just fit along the 10cm dimension.

Clearly you can fit 250 balls nicely into the box.

But then, if that were the answer, this wouldn't be much of a puzzle............

Innocent Bystander posted:

Innocent Bystander posted:

Don Atkinson posted:

Bill & Ben the flowerpot men

Two gardeners, Ben and Bill, are talking about their flowerpots.

“If you gave me one of your flowerpots”, said Ben, “I would have twice as many flowerpots as you”

“Well” said Bill, ”if, instead, you gave me one of yours, then I would have as many as you”

How many flowerpots does each gardener have ?

Too simple!!!  -But 2 possible answers

1) Ben 7 and Bill 5: Bill gives one makes 8,4 or Ben gives one makes 6,6

Can you see the second answer?

 

 

Now, this one has got me intrigued, so I am hoping others can help out ?

My arithmetic (algebra) doesn't suggest the possibility of a second set of numbers viz :-

Let X = Larger number

Let Y = Smaller number

X – 1 = Y + 1             (1)

X + 1 = 2(Y-1)           (2)

(1) ⇒  Y = X – 2          (3)

(2) ⇒ X + 1 = 2[(X – 2) – 1]  (ie Sub (3) into (2))

          X + 1 = 2X – 6

                X = 7          (4)

Sub (4) in (3)

            Y = 7 – 2

          Y = 5

Don Atkinson posted:

Division - aahhggg !!

Divide 45 into four parts such that the first part with two added, the second part with two subtracted, the third part multiplied by two and the fourth part divided by two, shall all be equal.

Thisshould  be a lot easier than arranging balls in boxes !

8, 12, 5, 20

solve in pairs of equal formulae. I first solved them all in tems of the smallest (3rd), finding 9y=45, and the rest simply fell into place 

Don Atkinson posted:

Innocent Bystander posted:

Innocent Bystander posted:

Don Atkinson posted:

Bill & Ben the flowerpot men

Two gardeners, Ben and Bill, are talking about their flowerpots.

“If you gave me one of your flowerpots”, said Ben, “I would have twice as many flowerpots as you”

“Well” said Bill, ”if, instead, you gave me one of yours, then I would have as many as you”

How many flowerpots does each gardener have ?

Too simple!!!  -But 2 possible answers

1) Ben 7 and Bill 5: Bill gives one makes 8,4 or Ben gives one makes 6,6

Can you see the second answer?

 

 

Now, this one has got me intrigued, so I am hoping others can help out ?

My arithmetic (algebra) doesn't suggest the possibility of a second set of numbers viz :-

Let X = Larger number

Let Y = Smaller number

X – 1 = Y + 1             (1)

X + 1 = 2(Y-1)           (2)

(1) ⇒  Y = X – 2          (3)

(2) ⇒ X + 1 = 2[(X – 2) – 1]  (ie Sub (3) into (2))

          X + 1 = 2X – 6

                X = 7          (4)

Sub (4) in (3)

            Y = 7 – 2

          Y = 5

Not sure I said it could be solved mathmatically....

Innocent Bystander posted:

Don Atkinson posted:

Innocent Bystander posted:

Innocent Bystander posted:

Don Atkinson posted:

Bill & Ben the flowerpot men

Two gardeners, Ben and Bill, are talking about their flowerpots.

“If you gave me one of your flowerpots”, said Ben, “I would have twice as many flowerpots as you”

“Well” said Bill, ”if, instead, you gave me one of yours, then I would have as many as you”

How many flowerpots does each gardener have ?

Too simple!!!  -But 2 possible answers

1) Ben 7 and Bill 5: Bill gives one makes 8,4 or Ben gives one makes 6,6

Can you see the second answer?

 

 

Now, this one has got me intrigued, so I am hoping others can help out ?

My arithmetic (algebra) doesn't suggest the possibility of a second set of numbers viz :-

Let X = Larger number

Let Y = Smaller number

X – 1 = Y + 1             (1)

X + 1 = 2(Y-1)           (2)

(1) ⇒  Y = X – 2          (3)

(2) ⇒ X + 1 = 2[(X – 2) – 1]  (ie Sub (3) into (2))

          X + 1 = 2X – 6

                X = 7          (4)

Sub (4) in (3)

            Y = 7 – 2

          Y = 5

Not sure I said it could be solved mathmatically....

I know you didn't.

I didn't obtain the 7 and 5 mathematically, just by inspection. I only ran through the algebra to see if it indicated any obscure solutions, And as you see, it didn't.

Hopefully others, with more perception,  are also looking for second answers.

Innocent Bystander posted:

Don Atkinson posted:

Division - aahhggg !!

Divide 45 into four parts such that the first part with two added, the second part with two subtracted, the third part multiplied by two and the fourth part divided by two, shall all be equal.

This should  be a lot easier than arranging balls in boxes !

8, 12, 5, 20

solve in pairs of equal formulae. I first solved them all in tems of the smallest (3rd), finding 9y=45, and the rest simply fell into place 

Nice rational approach.

I'm afraid my technique was more "Trial & Error". I started with the idea that "10" might be "the first part" and when that didn't work, tried "8" and much to my surprise.......

Regarding Bill and Ben, if the second answer is not a "maths"-based one, then perhaps it is something to do with how many flower pots the "real" Bill and Ben characters comprised of. I've looked at old pictures but unable to come up with a definitive answer.

........ or, semantically, the original question's title was Bill & Ben the flowerpot men (not flowerpots men), so maybe the second answer is that they each had a singular flowerpot? Just clutching at straws really!

steved posted:

........ or, semantically, the original question's title was Bill & Ben the flowerpot men (not flowerpots men), so maybe the second answer is that they each had a singular flowerpot? Just clutching at straws really!

 

Yes ! Hi Steve,

I've been wondering if its something along the lines of (say) Roman Numerals  eg XIX (19) take away I = XX (20)

Clue: ellipsis. (Not to be confused with ellipses!)

Don Atkinson posted:

Another Packing Problem

I have a box with internal dimensions 5cm x 5cm x 10cm and spherical balls of diameter 1cm. What is the maximum number of balls that I can fit into the box and close the lid?

Usual maths rules apply eg The balls are solid and won't squash. The box is solid and won't distort. All the dimensions are exact. ie 5 balls will just fit across the 5cm dimension, or 10 balls will just fit along the 10cm dimension.

By my calc, if 3D close packed (alternating layers of 5x5 & 4x4), 12 layers will fit in the 10cm height - but fit fewer than 250 in the box (246).   [Height between centres of rows is height of regular tetrahedron with edge length one diameter plus half a diameter each at top and bottom]

3D close packing the other way round with alternating layers of 10x5 and 9x4 won't fit even one extra layer compared to simple stacked packing, so a lot less in total.

Best I can fit is discreet hexagonally close packed layers as per the earlier cigarette example, fitting 12 rows alternating 5,4,5,4 etc in the 10cm dimension, giving a layer of 54  balls. A total of 5 such layers will fit in the box, so 270 in total.

Even discrete!

Innocent Bystander posted:

Even discrete!

Ah !

My English teacher once announced to the whole class, "Atkinson, you're not just the worst speller in the school at the moment, you are the worst speller this school has ever known !"

Praise indeed ! I'm sure the rest of the Forum considers you are an absolute dunce

With my pedigree, I shall be more discrete.

Given my post above, and my difficulty with English at school, you will appreciate that I might not be the best person here to sort out ellipsis, even if I have inadvertently created one !

But I shall have another look. Hopefully along with others.

Innocent Bystander posted:

Don Atkinson posted:

Another Packing Problem

I have a box with internal dimensions 5cm x 5cm x 10cm and spherical balls of diameter 1cm. What is the maximum number of balls that I can fit into the box and close the lid?

Usual maths rules apply eg The balls are solid and won't squash. The box is solid and won't distort. All the dimensions are exact. ie 5 balls will just fit across the 5cm dimension, or 10 balls will just fit along the 10cm dimension.

By my calc, if 3D close packed (alternating layers of 5x5 & 4x4), 12 layers will fit in the 10cm height - but fit fewer than 250 in the box (246).   [Height between centres of rows is height of regular tetrahedron with edge length one diameter plus half a diameter each at top and bottom]

3D close packing the other way round with alternating layers of 10x5 and 9x4 won't fit even one extra layer compared to simple stacked packing, so a lot less in total.

Best I can fit is discreet hexagonally close packed layers as per the earlier cigarette example, fitting 12 rows alternating 5,4,5,4 etc in the 10cm dimension, giving a layer of 54  balls. A total of 5 such layers will fit in the box, so 270 in total.

Well IB, that's a jolly good start, and i'm pleased someone has the real courage to get the ball rolling, so to speak (sorry !)

You might have gathered by now that more than 270 balls can be fitted into this box. Spheres 1cm dia, box 5x5x10 and it doesn't require ellipsis (or ellipses), just slightly broader thinking and arrangements. And you clearly have the right sort of arithmetic, or geometry available to cope.

Just a short quick one to keep the balls rolling..........

Easy  North American currency (rather than £-s-d)

Twenty years ago I first helped to set up an enterprise in BC Canada.

One day I popped into a store and spent half the money I had in my pocket.

On leaving the store, I realised that I now had as many cents in my pocket as I had previously had dollars and I also had half as many dollars as I previously had had cents.

How much did I have when I entered to store ?

PS. Twenty years ago cents were in use, I appreciate that now we are stuck at 5 cent pieces as the smallest coin in use.

PPS. No ellipsis ( or even ellipses allowed !!)

Don Atkinson posted:

Innocent Bystander posted:

Don Atkinson posted:

Another Packing Problem

I have a box with internal dimensions 5cm x 5cm x 10cm and spherical balls of diameter 1cm. What is the maximum number of balls that I can fit into the box and close the lid?

Usual maths rules apply eg The balls are solid and won't squash. The box is solid and won't distort. All the dimensions are exact. ie 5 balls will just fit across the 5cm dimension, or 10 balls will just fit along the 10cm dimension.

By my calc, if 3D close packed (alternating layers of 5x5 & 4x4), 12 layers will fit in the 10cm height - but fit fewer than 250 in the box (246).   [Height between centres of rows is height of regular tetrahedron with edge length one diameter plus half a diameter each at top and bottom]

3D close packing the other way round with alternating layers of 10x5 and 9x4 won't fit even one extra layer compared to simple stacked packing, so a lot less in total.

Best I can fit is discreet hexagonally close packed layers as per the earlier cigarette example, fitting 12 rows alternating 5,4,5,4 etc in the 10cm dimension, giving a layer of 54  balls. A total of 5 such layers will fit in the box, so 270 in total.

Well IB, that's a jolly good start, and i'm pleased someone has the real courage to get the ball rolling, so to speak (sorry !)

You might have gathered by now that more than 270 balls can be fitted into this box. Spheres 1cm dia, box 5x5x10 and it doesn't require ellipsis (or ellipses), just slightly broader thinking and arrangements. And you clearly have the right sort of arithmetic, or geometry available to cope.

I will provide the answer ie the number of 1cm dia spheres that can be fitted into the box tomorrow (unless somebody comes up with the answer before then).

All that will then be required is the "how do we fit that many into the box"

PS, it's not an ellipsis, just plane, simple, geometry !

Mu-so Qb at Two-a-Penny

(another topical quickie to accompany the Dollar Quest above, whilst we come to grips with the 1cm spheres)

I noticed in another thread (in the HiFi Corner) that Naim Mu-so Qb are now two-a-penny……..(Black Friday in the UK only)

Actually, a bit of research shows they are being sold at three prices depending on whether they are Brand New, Ex-Display or Refurbished viz respectively one a penny, two a penny and three a penny.

A group of Forum Members (there were as many men as women) had between them the grand total of sevenpence to spend on these Mu-so Qb, each receiving exactly alike (and in doing so spending the precise sevenpence).

How many Mu-so Qb, and of which version, did each receive ?

Don Atkinson posted:

Don Atkinson posted:

Innocent Bystander posted:

Don Atkinson posted:

Another Packing Problem

I have a box with internal dimensions 5cm x 5cm x 10cm and spherical balls of diameter 1cm. What is the maximum number of balls that I can fit into the box and close the lid?

Usual maths rules apply eg The balls are solid and won't squash. The box is solid and won't distort. All the dimensions are exact. ie 5 balls will just fit across the 5cm dimension, or 10 balls will just fit along the 10cm dimension.

By my calc, if 3D close packed (alternating layers of 5x5 & 4x4), 12 layers will fit in the 10cm height - but fit fewer than 250 in the box (246).   [Height between centres of rows is height of regular tetrahedron with edge length one diameter plus half a diameter each at top and bottom]

3D close packing the other way round with alternating layers of 10x5 and 9x4 won't fit even one extra layer compared to simple stacked packing, so a lot less in total.

Best I can fit is discreet hexagonally close packed layers as per the earlier cigarette example, fitting 12 rows alternating 5,4,5,4 etc in the 10cm dimension, giving a layer of 54  balls. A total of 5 such layers will fit in the box, so 270 in total.

Well IB, that's a jolly good start, and i'm pleased someone has the real courage to get the ball rolling, so to speak (sorry !)

You might have gathered by now that more than 270 balls can be fitted into this box. Spheres 1cm dia, box 5x5x10 and it doesn't require ellipsis (or ellipses), just slightly broader thinking and arrangements. And you clearly have the right sort of arithmetic, or geometry available to cope.

I will provide the answer ie the number of 1cm dia spheres that can be fitted into the box tomorrow (unless somebody comes up with the answer before then).

All that will then be required is the "how do we fit that many into the box"

PS, it's not an ellipsis, just plane, simple, geometry !

One thought I had was expanding on my 270 solution, inverting the alternating layers so 4.5.4 overlays 5,4,5, but altough occupying less space still only 5 layers would fit but the free space of very nearly one layer does suggest more could fit. The only thing left is spacing wider than all balls touching in a layer, allowing the next later to sit lower - but having been away this weekend I haven't had time to determine what would be the right calculation. 

Innocent Bystander posted:

One thought I had was expanding on my 270 solution, inverting the alternating layers so 4.5.4 overlays 5,4,5, but altough occupying less space still only 5 layers would fit but the free space of very nearly one layer does suggest more could fit. The only thing left is spacing wider than all balls touching in a layer, allowing the next later to sit lower - but having been away this weekend I haven't had time to determine what would be the right calculation. 

 

I like your thinking................

By “I like your thinking” I think you are heading in the right sort of direction.

cheers Don

Got it: 297.

actually simpler than I was thinking it would be! Basis of calcs below as already given for simple hexagonal packing of a layer:

Layers = 11 rows 5,4,5...5, alternating with 4,5,4...4. Sitting directly on top gives row height between centres sqrt 0.75 (~0.866), and only 5 layers fit*.

The spare space at end of each layer is 10-(~0.866x10 + 0.5 x2) = 0.33975, by up to which amount the 2nd, 4th and 6th layers can be staggered. As this is less than half the 0.866 spacing of row centres the staggering can be by this full amount without the height starting to go up again.
So as the second layer is shifted simple pythagorus gives the height between row centres: hypotaneuse is sqrt 0.75, base is 0.33975 so height between centres is 0.7966
6 layers = 5x this + 2x0.5 = 4.98, so fits in height of 5.
That is 6 layers of 11 rows 5,4,5...5 alternating 4,5,4...4, = 50x3 + 49x3 =297

*There was actually a mistake in my previous calc of 270, as that was 12 rows per layer 5,4,5...4, but at 11x0.866 between centres +2x0.5 that is more than 10 (for some reason I picked the wrong figure in my working. So with only 11 rows fitting, max would be 5,4,5...5 which is 50 per layer and 5 layers so 250 (not 270 I stated) - the same as simple packing but space at end.

BC Canada Enterprise - I think Don had 99 dollars and 98 cents in his pocket to start with.

Please could Innocent Bystander put us out of our misery regarding Bill & Ben!