A Fistful of Brain Teasers

steved posted:

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!

The BC Canada answer appears to fit.

No takers on the elliptic one?

steved posted:

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!

You think right, Steve. Neat.

And well supported by IB, nice.

Innocent Bystander posted:

steved posted:

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!

The BC Canada answer appears to fit.

No takers on the elliptic one?

Looks like both Steve and myself (and I guess a few “browsers”) are pretty well stumped over this elusive elliptic and need early enlightenment...... .......

cheers, Don

OK

Ellipsis already exists in the problem as given: When Bill and Ben say: "I would have ... as you", they perhaps appear to mean: "I would have ... as you [would have after the transaction]". That was my first reading, leading me to deduce that Bill has 5 and Ben 7.

However the words equally could mean: "I would have ... as you [have now]". So, Bill has 2 and Ben 3. (Bill gives one so Ben has 4, twice Bill's original 2, or Ben gives one and Bill has 3, the same as Ben's original quantity.)

Potentially there could be two more possible answers: Ben could mean "...as you [would have after the transaction]" with Bill meaning "...as you [have now]", giving answers of Bill has 4 and Ben 5, or vice versa giving answers of Bill 3 and Ben 5 - however given that they both clearly know the number the other has, a different implied meaning would be deliberately obtuse of Bill (the person who spoke second).

Innocent Bystander posted:

OK

Ellipsis already exists in the problem as given: When Bill and Ben say: "I would have ... as you", they perhaps appear to mean: "I would have ... as you [would have after the transaction]". That was my first reading, leading me to deduce that Bill has 5 and Ben 7.

However the words equally could mean: "I would have ... as you [have now]". So, Bill has 2 and Ben 3. (Bill gives one so Ben has 4, twice Bill's original 2, or Ben gives one and Bill has 3, the same as Ben's original quantity.)

Potentially there could be two more possible answers: Ben could mean "...as you [would have after the transaction]" with Bill meaning "...as you [have now]", giving answers of Bill has 4 and Ben 5, or vice versa giving answers of Bill 3 and Ben 5 - however given that they both clearly know the number the other has, a different implied meaning would be deliberately obtuse of Bill (the person who spoke second).

I’m going to have to read this a few times just to capture the sequence of words, never mind the meaning......

nice one !

The greatest number of spheres in the box is 288.

Innocent Bystander posted:

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.

I’m still trying to visualise the layers and rows and their relative positions as described above. 

288 might be wrong, but the original author (not me) was pretty convincing, and my solution matched his.

Perhaps you could re-describe your solution using the word “layer” to define a group of spheres with their centres in a common plane, and stating how many spheres are in each layer, and the layout of spheres within the layer.

cheers, Don

Don Atkinson posted:

Innocent Bystander posted:

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.

I’m still trying to visualise the layers and rows and their relative positions as described above. 

288 might be wrong, but the original author (not me) was pretty convincing, and my solution matched his.

Perhaps you could re-describe your solution using the word “layer” to define a group of spheres with their centres in a common plane, and stating how many spheres are in each layer, and the layout of spheres within the layer.

cheers, Don

 

Will do, as clearly as I can!

Many thanks, you could well have a superb solution....hopefully, we can confirm it !

Don, regarding the Muso puzzle, unless I've misunderstood the question, there is more than answer as far as I can tell:-

There are 2 forum members (1 male, 1 female).

Each buys 1 Brand New (1p) , 1 Exdemo (0.5p) and 6 Reconditioned (2p)

or Each buys 2 Brand new (2p), 1 Exdemo (0.5p) and 3 Reconditioned (1p)

or Each buys 1 Brand New (1p), 3 Exdemo (1.5p) and 3 Reconditioned (1p)

Am I missing something?

steved posted:

Don, regarding the Muso puzzle, unless I've misunderstood the question, there is more than answer as far as I can tell:-

There are 2 forum members (1 male, 1 female).

Each buys 1 Brand New (1p) , 1 Exdemo (0.5p) and 6 Reconditioned (2p)

or Each buys 2 Brand new (2p), 1 Exdemo (0.5p) and 3 Reconditioned (1p)

or Each buys 1 Brand New (1p), 3 Exdemo (1.5p) and 3 Reconditioned (1p)

Am I missing something?

 

I think the words “a group of.....” was meant to signify “more than two”. 

But I have this horrible feeling this might lead to more “ellipsis”

Where’s IB when you need him!

Sorry for the confusion Steve, and hope this makes it a slightly more interesting challenge.

The answer I got was 3 men and 3 women. Each getting 2 refurbished and 1 ex-demo. I hope there aren’t too many more correct answers !

Don Atkinson posted:

steved posted:

Don, regarding the Muso puzzle, unless I've misunderstood the question, there is more than answer as far as I can tell:-

There are 2 forum members (1 male, 1 female).

Each buys 1 Brand New (1p) , 1 Exdemo (0.5p) and 6 Reconditioned (2p)

or Each buys 2 Brand new (2p), 1 Exdemo (0.5p) and 3 Reconditioned (1p)

or Each buys 1 Brand New (1p), 3 Exdemo (1.5p) and 3 Reconditioned (1p)

Am I missing something?

 

I think the words “a group of.....” was meant to signify “more than two”. 

But I have this horrible feeling this might lead to more “ellipsis”

Where’s IB when you need him!

Sorry for the confusion Steve, and hope this makes it a slightly more interesting challenge.

The answer I got was 3 men and 3 women. Each getting 2 refurbished and 1 ex-demo. I hope there aren’t too many more correct answers !

Quick look in dictionary defines a group as 2 or more.

Yes, English meaning, but not ellipsis which is a meaning that is omitted from a sentence as being understood (usually unambiguously as in "Come here" meaning "[You] come here", the word 'you' being understood by all.

Don Atkinson posted:

Innocent Bystander posted:

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.

I’m still trying to visualise the layers and rows and their relative positions as described above. 

288 might be wrong, but the original author (not me) was pretty convincing, and my solution matched his.

Perhaps you could re-describe your solution using the word “layer” to define a group of spheres with their centres in a common plane, and stating how many spheres are in each layer, and the layout of spheres within the layer.

cheers, Don

 

Ok, here goes -I hope this is clear, as it is hard to be sure someone else will picture what I am trying to describe in words:

Box sits on a long (10*5) side, and for ease has sides and ends, open at top.

First layer 5,4,5,4,5,4,5,4,5,4,5 =11 rows (50 balls).
Hexagonal close packing, actually could be called triangular close packing, ball centres on equilateral triangles, sides 1diameter (1cm), spacing between row centres is height of the triangle: using pythag = sqrt 0.75 (~0.866)

Total length of 11 rows = 10*~0.866 + 2*0.5 = ~9.66. Spare space at end ~0.34

Second layer 4,5,4,5,4,5,4,5,4,5,4 =11 rows (49 balls).
Spacing exactly as first layer.

3rd-6th layers repeating 1&2.

Starting point is all vertically aligned at, say, left hand end of box, so clear 0.34cm end space at right hand end.

At this point there are vertical rows, 1st starting at bottom left: 5,4,5,4,5,4 2nd vertical row 4,5,4,5,4,5 etc.

Spacing between vertical row centres at starting position (left aligned) is same as horizontal rows = sqrt 0.75 (~0.866). Starting height = 0.866*4 + 2*0.5 = 5.33 (so proud of the top of box)

Now push layers 2,4 & 6 horizontally to the right by 0.33975 cm to touch the right hand end of the box, layers 1,3,5 remaining against left end.. The layers drop: Vertical layer spacing starts at 0.866 as described above. The equilateral triangle of 2 balls bottom layer 1 ball second stays intact as the upper ball moves and lowers, forming the hypotaneuse of a right angled triangle, its height being the height of the upper ball centre above the lower ball centre, and base being the amount the upper row is displaced when pushed. Vertical height (layer spacing) after moving 0.33975cm to the right = 0.7966 (pythag)

6 layers with centres spaced by 0.7966 gives total height 0.7966*5 + 2*0.5 = 4.983, which does fit in box.

No. Balls = 3x50 + 3*49 = 397.

The only problem would be if the balls in the second layer were to contact the next row balls of the first layer before the second layer was fully shifted by 0.33975. However, when I originally did this I didn't think they would contact until the 2nd layer balls had moved by 0.433 (half the horizontal row spacing of 0.866), which was more than in this case so ok... BUT I now realise I was wrong and the lowest point before contacting the next row ball would be the cevtre point of the equilateral triangle of the lower layer with base on the centre points of row 1 balls, so distance to triangle centre is given by half base (0.5cm)*tan 30 =0.28868. This would be max drop in height, and is less than with the 0.33975 shift calculated above. Spacing of layer centres at 0.28868 shift (pythag calc as above with hypot. length 0.866) is 0.8164 so total height of 6 layers exceeds 5 (5*0.8164 +2*0.5) - in fact this is back to full 3D close packing I had already found didn't work! 

So my answer was wrong. The question is then will 288 work, and how? My first thought is space rows 1,3 & 5 by 0.03395 (1/10th of the excess 0.3395), and omit the last row of 4 in rows 2,4,6, which would need a new calc for the amount of lowering (as I had previously postulated, but not quite as simple to calc as the layer packing is no longer regular hexagon/triangle) - however that would be a reduction of 12 from the 297 above, i.e 285, so doesn't fit with 288. 

Hmm, have to cone back to it.

Innocent Bystander posted:

Don Atkinson posted:

steved posted:

Don, regarding the Muso puzzle, unless I've misunderstood the question, there is more than answer as far as I can tell:-

There are 2 forum members (1 male, 1 female).

Each buys 1 Brand New (1p) , 1 Exdemo (0.5p) and 6 Reconditioned (2p)

or Each buys 2 Brand new (2p), 1 Exdemo (0.5p) and 3 Reconditioned (1p)

or Each buys 1 Brand New (1p), 3 Exdemo (1.5p) and 3 Reconditioned (1p)

Am I missing something?

 

I think the words “a group of.....” was meant to signify “more than two”. 

But I have this horrible feeling this might lead to more “ellipsis”

Where’s IB when you need him!

Sorry for the confusion Steve, and hope this makes it a slightly more interesting challenge.

The answer I got was 3 men and 3 women. Each getting 2 refurbished and 1 ex-demo. I hope there aren’t too many more correct answers !

Quick look in dictionary defines a group as 2 or more.

Yes, English meaning, but not ellipsis which is a meaning that is omitted from a sentence as being understood (usually unambiguously as in "Come here" meaning "[You] come here", the word 'you' being understood by all.

Oh dear. English never was my strong subject.

Non-the-less, Steve did come up with perfectly valid answers, even though he wondered why the "problem" was so "easy" and with so many possible solutions.

Well done Steve, and I will try to be more careful with my wording in future.

PS, a Brain Teaser last week on that Fishy Forum ran to five pages or more disputing the wording. So many thanks guys for limiting my "walk of shame" here to a couple of polite posts !

Innocent Bystander posted:

 

Ok, here goes -I hope this is clear, as it is hard to be sure someone else will picture what I am trying to describe in words:Nicely described and very clear down to "= 397"  (which should read "= 297" btw)

Box sits on a long (10*5) side, and for ease has sides and ends, open at top.

First layer 5,4,5,4,5,4,5,4,5,4,5 =11 rows (50 balls).
Hexagonal close packing, actually could be called triangular close packing, ball centres on equilateral triangles, sides 1diameter (1cm), spacing between row centres is height of the triangle: using pythag = sqrt 0.75 (~0.866)

Total length of 11 rows = 10*~0.866 + 2*0.5 = ~9.66. Spare space at end ~0.34

Second layer 4,5,4,5,4,5,4,5,4,5,4 =11 rows (49 balls).
Spacing exactly as first layer.

3rd-6th layers repeating 1&2.

Starting point is all vertically aligned at, say, left hand end of box, so clear 0.34cm end space at right hand end.

At this point there are vertical rows, 1st starting at bottom left: 5,4,5,4,5,4 2nd vertical row 4,5,4,5,4,5 etc.

Spacing between vertical row centres at starting position (left aligned) is same as horizontal rows = sqrt 0.75 (~0.866). Starting height = 0.866*4 + 2*0.5 = 5.33 (so proud of the top of box)

Now push layers 2,4 & 6 horizontally to the right by 0.33975 cm to touch the right hand end of the box, layers 1,3,5 remaining against left end.. The layers drop: Vertical layer spacing starts at 0.866 as described above. The equilateral triangle of 2 balls bottom layer 1 ball second stays intact as the upper ball moves and lowers, forming the hypotaneuse of a right angled triangle, its height being the height of the upper ball centre above the lower ball centre, and base being the amount the upper row is displaced when pushed. Vertical height (layer spacing) after moving 0.33975cm to the right = 0.7966 (pythag)

6 layers with centres spaced by 0.7966 gives total height 0.7966*5 + 2*0.5 = 4.983, which does fit in box.

No. Balls = 3x50 + 3*49 = 397.

This next bit is not so clear to me, but it has helped you realise your 297 solution doesn't work, which is a big step forward.

The only problem would be if the balls in the second layer were to contact the next row balls of the first layer before the second layer was fully shifted by 0.33975. However, when I originally did this I didn't think they would contact until the 2nd layer balls had moved by 0.433 (half the horizontal row spacing of 0.866), which was more than in this case so ok... BUT I now realise I was wrong and the lowest point before contacting the next row ball would be the cevtre point of the equilateral triangle of the lower layer with base on the centre points of row 1 balls, so distance to triangle centre is given by half base (0.5cm)*tan 30 =0.28868. This would be max drop in height, and is less than with the 0.33975 shift calculated above. Spacing of layer centres at 0.28868 shift (pythag calc as above with hypot. length 0.866) is 0.8164 so total height of 6 layers exceeds 5 (5*0.8164 +2*0.5) - in fact this is back to full 3D close packing I had already found didn't work! 

This next bit is beginning to look promising again

So my answer was wrong. The question is then will 288 work, and how? My first thought is space rows 1,3 & 5 by 0.03395 (1/10th of the excess 0.3395), and omit the last row of 4 in rows 2,4,6, which would need a new calc for the amount of lowering (as I had previously postulated, but not quite as simple to calc as the layer packing is no longer regular hexagon/triangle) - however that would be a reduction of 12 from the 297 above, i.e 285, so doesn't fit with 288. 

What would happen if you allowed the first, third and fifth layers to "expand" evenly along the base of your box ? then rows 2, 4, 6 would "sink" a little lower between the spheres of rows 1,3 and 5.   I don't think this will get your 297 into the box but it might help you (and others who might be watching from the wings) to think "outside the box" (sorry for the pun - no spheres end up outside the box !)

Hmm, have to cone back to it.

 

 

You might try turning the box on end, so that it resembles a box containing a bottle of wine or champagne. ie one 5x5 dimension forms the base and the four 10x5 dimensions form the vertical sides. The second 5x5 dimension forms the top. This is only a suggestion, but it enabled me to visualise my solution more easily. There re no "tricks" or ellipsis in this problem, just a bit of geometrical imagination and arithmetic (trig or geometry).

Hi Don, thanks for the clarification. For my part, given that you had included 3 products, I assumed that a solution had to include all three, and two people was the only way I could find to achieve that. Another "correct" solution would be that 14 people all bought just one Ex-display each.

My closing thought above didn't work, so wth the clue of 288 I started some reasoning. Tall way round layers must be <25 balls per layer as I have already eliminated that. 288 is 18 layers or other repeats of 16, (or 16 of 18), or 24 of 12.. I did hit on what I thought worked perfectly: evenly spaced 4*4 full out to sides (container vertical), next layer 3*3 in the gaps, repeat, but no, only 241. Or 4*4 spaced by 0.1428 from one corner, alternating layer same spacing from opposite corner ferfectly meshing - but only 16 layers of 16 = 256. Something similar lying on long side 7 layers of 4x9 = 251. So, in conclusion The best I have actually found so far is 256. Sadly not 288. I think I'm running out of ideas...

Innocent Bystander posted:

My closing thought above didn't work, so wth the clue of 288 I started some reasoning. Tall way round layers must be <25 balls per layer as I have already eliminated that.288 is 18 layers or other repeats of 16, (or 16 of 18), or 24 of 12.. I did hit on what I thought worked perfectly: evenly spaced 4*4 full out to sides (container vertical), next layer3*3 in the gaps, repeat, but no, only 241. Or 4*4 spaced by 0.1428 from one corner, alternating layer same spacing from opposite corner ferfectly meshing - but only 16 layers of 16 = 256. Something similar lying on long side 7 layers of 4x9 = 251. So, in conclusion The best I have actually found so far is 256. Sadly not 288. I think I'm running out of ideas...

You are sooo.... close ! concentrate on your idea of the 16 layers of 18. ie 18 balls in each of 16 layers.

(it's much better if you can find or even stumble upon the solution rather than simply be told it, even if it means more hints !)

Indeed (and I only want to be told the solution in order to be able to check and (maybe!) prove wrong! I,ll have another look later this eve or tomorrow ...unless someone beats me to it

steved posted:

Hi Don, thanks for the clarification. For my part, given that you had included 3 products, I assumed that a solution had to include all three, and two people was the only way I could find to achieve that. Another "correct" solution would be that 14 people all bought just one Ex-display each.

OMG.....

it seems there are more correct solutions to this one than wrong answers !!!!

Nice one Steve !

Now, as far as I know, this one has only one solution. ( I can't guarantee that, but I do believe it is the case)  And I hope (and pray) that I haven't muddled up the wording to confuse Steve (or anybody else) or allow IB to spot an ellipsis....

Animal Farm (fingers crossed)

A farmer went to market and bought a hundred animals at a total cost of £100. The price of cows was £5 each, Sheep £1 each and chickens £1 each.

How many of each kind did he buy ?

Animal Farm:-

Hi Don. Erm..... Are those animal prices correct? If so, there are lots of "correct" permutations involving Sheep and Chickens, but no Cows.

steved posted:

Animal Farm:-

Hi Don. Erm..... Are those animal prices correct? If so, there are lots of "correct" permutations involving Sheep and Chickens, but no Cows.

Omg no, not again !

i’m so Sorry. I’ ll Have to read my notes again and get back to you tomorrow.

Thank you for effectively proof reading my rubbish yet again.  I must be cracking up !

And I'm guessing the question is missing some info about the types of animal the farmer actually buys (including not omitting any animal type)?

Innocent Bystander posted:

And I'm guessing the question is missing some info about the types of animal the farmer actually buys (including not omitting any animal type)?

You're guessing right.......!