A Fistful of Brain Teasers

Peakman posted:

Hi Don

I think the distance is 200 miles and the train started at 50mph.

Only 2 hours late sounds pretty good by our local standards.

Roger

Hi Roger,

Looks like you are on "Peak" performance with the old maths !!

200 miles and 50mph are spot-on.

Surprisingly, out trains from Newbury to Paddington are very punctual. Milchester is some sort of exception.....I've never heard of a train actually arriving there yet

Another one for the Hikers and Bikers.........

Myself and Mrs D have to share our one and only bicycle. (it’s not a true storey !)

We often pop into Reading for a spot of lunch and share the bike to make the journey quicker. We like to leave home at the same time and arrive in Reading at the same time. The distance is 20 miles.

Mrs D can only walk at 4 mph whilst I can walk at 5 mph. But Mrs D can ride at 10 mph whilst I can only manage 8 mph (how I wish winky was ready to sell his new bike for a few peanuts !)

How do we arrange the journey ?

You can assume we walk and ride at steady speeds without rests and no time is lost getting on/off the bike etc etc.

Eoink posted:

Ah, in that case, I make it 21mph. (And thanks for the 135 in the question, much appreciated.)

Hi Eoink,

that's right, 21 mph.

and the 135 was the closest I could get to your (the thought being sparked by the 62 in the one above). So 12, 22, 32, 32.5..., 90, 110, 135...) challenge !

Nick from Suffolk posted:

24.5mph, but I have finished my proof reading and drunk nearly 2 bottles of wine, at a cost great than the MAP proposed limit. I will leave you to ascertain which is more wrong.

Ah Nick,

Let's put the 24.5 mph (vi 21 mph) "near-miss" down to my  poor wording, rather than your 2 bottles of wine and proof-reading.....

Don Atkinson posted:

Another one for the Hikers and Bikers.........

Myself and Mrs D have to share our one and only bicycle. (it’s not a true storey !)

We often pop into Reading for a spot of lunch and share the bike to make the journey quicker. We like to leave home at the same time and arrive in Reading at the same time. The distance is 20 miles.

Mrs D can only walk at 4 mph whilst I can walk at 5 mph. But Mrs D can ride at 10 mph whilst I can only manage 8 mph (how I wish winky was ready to sell his new bike for a few peanuts !)

How do we arrange the journey ?

You can assume we walk and ride at steady speeds without rests and no time is lost getting on/off the bike etc etc.

Mrs D starts on the bike, accelarates instantly to a precise 10mph, and after precisely 66 minutes 40 secs she stops instantly, managing to avoid somersaulting over the handlebars. And starts walking, at a precise 4mph. You set out at the same time, walking at exactly 5mph, and after 2 hours 13 mins start looking ahead for the bike, hoping it hasn't been nicked as she didn't have time to lock it. 20 seconds later you reach it, mount and accelerate instantly to a precise 8mph (you really need to get out cycling more, to get fitter). 66 minutes and 40 secs later you collide with your poor wife, having just swerved onto the pavement to avoid a maniac in a car. Hopefully that is in Castle St. right outside Sweeney & Todd, the delightful pub-restaurant where you can recover over a delicious home-made pie and rattatoille washed down by a nice pint or two of real ale (hoping it is still there - it is many years since I lived there).

Sweeney and Todd is still there and doing well. Still serving really good pies and ale. I guess it must be about 30 years since I first went there. At the time I was providing a consultancy service to an engineering firm based on the far side of town where Sutton Seeds had once been.

Oh ! and your brain teaser timings were precise and correct. Brilliant.

PS I think if I can accelerate from 0 to 8 mphinstantly, then i'm fit enough.....almost as good as winky's Tesla

I wish I could do a brain teaser, unfortunately I seem more skilled at intervening in happy memories that have caused a smile....Sweeney and Todd is up for sale so if you want to enjoy the current ownership be swift....http://www.getreading.co.uk/news/business/reading-sweeney-todd-for-sale-13816425

Sorry to hear that - but then my frequenting of it was also in mid 80s, and last time i was therevwas visiting in maybe late 90s, so it is good to know it has survived this long.

It’s up for sale, but hopefully not closing.

Probably not such a bad idea, one or two people I know, have said it needs to rejuvenate.

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 ?

Musing about a Cube..........or should that be Mu-so Qb

The numbers 370 and 407 each have the peculiar feature that the sum of the cubes of their digits is equal to the number itself.

For example, Cube of 3 (3^3) = 27; Cube of 7 = 343; 0^3 = 0

27 + 343 + 0 = 370

Can you find another number with this peculiarity, but NOT containing a “nought”

We will exclude the rather trivial number “1”

153 and 371.

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?

Nothing like as complicated (or elegant) as SteveD's offering from way-back-when, but how many more cigarettes could be packed into the same carton

Am I missing something? Surely the answer can’t be 135?

Hungryhalibut posted:

Am I missing something? Surely the answer can’t be 135?

The carton currently contains 160 cigarettes. I haven’t drawn all 160, just some of them to illustrate how they are packed ie 20 rows, each stacked eight cigarettes deep.

It should be possible to repack the carton differently and thus pack more cigarettes into the carton.

Hope this helps to make things more clear ?

151

20 and 19 alternating fits one more row of 20 on top, 176 total.

Hungryhalibut posted:

153 and 371.

Neat !!

Innocent Bystander posted:

151

20 and 19 alternating fits one more row of 20 on top, 176 total.

Not sure what the 151 represents ?

176 is certainly possible with the cigarettes. Anybody want to provide the arithmetic to support this ?

Don Atkinson posted:

Innocent Bystander posted:

151

20 and 19 alternating fits one more row of 20 on top, 176 total.

Not sure what the 151 represents ?

176 is certainly possible with the cigarettes. Anybody want to provide the arithmetic to support this ?

151 is the remaining number that can be fitted as per the question (how many moter...), there being 25 in there already.

Arithmetic is the centres of the circles that represent the cigs are at the points of regular hexagons, with a flat side horizontal  - picture as two circles at bottom, three centred on top and two above (if there were drawing tools here it would be so much easier!). Hexagon side length is one cig diameter. Height between two rows is half the distance between two parallel sides of hexagon (calc from length of side using pythagorus) =0.866 diameters. Divide total height (8 diameters) by this =9.238 rows, so 9 in practice. That is 4 double rows of 20 with 19 on top, plus an additional row of 20. Whether another one or two more can fit in by tilting and shaking about I cannot estimate.

Innocent Bystander posted:

Don Atkinson posted:

Innocent Bystander posted:

151

20 and 19 alternating fits one more row of 20 on top, 176 total.

Not sure what the 151 represents ?

176 is certainly possible with the cigarettes. Anybody want to provide the arithmetic to support this ?

151 is the remaining number that can be fitted as per the question (how many moter...), there being 25 in there already.

Arithmetic is the centres of the circles that represent the cigs are at the points of regular hexagons, with a flat side horizontal  - picture as two circles at bottom, three centred on top and two above (if there were drawing tools here it would be so much easier!). Hexagon side length is one cig diameter. Height between two rows is half the distance between two parallel sides of hexagon (calc from length of side using pythagorus) =0.866 diameters. Divide total height (8 diameters) by this =9.238 rows, so 9 in practice. That is 4 double rows of 20 with 19 on top, plus an additional row of 20. Whether another one or two more can fit in by tilting and shaking about I cannot estimate.

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.

Don Atkinson posted:

Innocent Bystander posted:

Don Atkinson posted:

Innocent Bystander posted:

151

20 and 19 alternating fits one more row of 20 on top, 176 total.

Not sure what the 151 represents ?

176 is certainly possible with the cigarettes. Anybody want to provide the arithmetic to support this ?

151 is the remaining number that can be fitted as per the question (how many moter...), there being 25 in there already.

Arithmetic is the centres of the circles that represent the cigs are at the points of regular hexagons, with a flat side horizontal  - picture as two circles at bottom, three centred on top and two above (if there were drawing tools here it would be so much easier!). Hexagon side length is one cig diameter. Height between two rows is half the distance between two parallel sides of hexagon (calc from length of side using pythagorus) =0.866 diameters. Divide total height (8 diameters) by this =9.238 rows, so 9 in practice. That is 4 double rows of 20 with 19 on top, plus an additional row of 20. Whether another one or two more can fit in by tilting and shaking about I cannot estimate.

Ah ! Got it.

My response to HH explains why I only drew 25 cigarettes - it was lazyness on my part. I had hoped the "160 cigarettes" caption made it clear how many cigarettes were initially packed into the carton.

One of these days I might get around to looking at 19 cigarettes on the bottom row, 18 in the next layer, then 19 etc thus giving 185, which might, or might not fit........one of these days............

Meanwhile, 176 is as good as I've seen, and with nice arithmetic.

Cheers

Don

I believe hexagonal close packing which this is (as in beehives) is the tightest possibe ...but that might not be the maximum possible if there is dead space at the top, as there is a tiny bit with this:  however19, 18, 19 would only fit 9 rows into a height of 8 units, so 167 in all. The case would need to be 8.644 units tall to fit a 10th row and so contain 185.

Incidentally, there was an error in my maths stated above for 20, 19, 20 packing: should be 0.866 diameters between centre of bottom and centre each row above up to centre of top rows + 2x 0.5 diameter for the bottom half of bottom row and top half of top row - but still fits (7.928 units high)

Innocent Bystander posted:

Don Atkinson posted:

Innocent Bystander posted:

Don Atkinson posted:

Innocent Bystander posted:

151

20 and 19 alternating fits one more row of 20 on top, 176 total.

Not sure what the 151 represents ?

176 is certainly possible with the cigarettes. Anybody want to provide the arithmetic to support this ?

151 is the remaining number that can be fitted as per the question (how many moter...), there being 25 in there already.

Arithmetic is the centres of the circles that represent the cigs are at the points of regular hexagons, with a flat side horizontal  - picture as two circles at bottom, three centred on top and two above (if there were drawing tools here it would be so much easier!). Hexagon side length is one cig diameter. Height between two rows is half the distance between two parallel sides of hexagon (calc from length of side using pythagorus) =0.866 diameters. Divide total height (8 diameters) by this =9.238 rows, so 9 in practice. That is 4 double rows of 20 with 19 on top, plus an additional row of 20. Whether another one or two more can fit in by tilting and shaking about I cannot estimate.

Ah ! Got it.

My response to HH explains why I only drew 25 cigarettes - it was lazyness on my part. I had hoped the "160 cigarettes" caption made it clear how many cigarettes were initially packed into the carton.

One of these days I might get around to looking at 19 cigarettes on the bottom row, 18 in the next layer, then 19 etc thus giving 185, which might, or might not fit........one of these days............

Meanwhile, 176 is as good as I've seen, and with nice arithmetic.

Cheers

Don

I believe hexagonal close packing which this is (as in beehives) is the tightest possibe ...but that might not be the maximum possible if there is dead space at the top, as there is a tiny bit with this:  however19, 18, 19 would only fit 9 rows into a height of 8 units, so 167 in all. The case would need to be 8.644 units tall to fit a 10th row and so contain 185.

Incidentally, there was an error in my maths stated above for 20, 19, 20 packing: should be 0.866 diameters between centre of bottom and centre each row above up to centre of top rows + 2x 0.5 diameter for the bottom half of bottom row and top half of top row - but still fits (7.928 units high)

One of these days....

....was really intended as a hint for the 1cm spheres into the 5x5x10 box teaser.......but

you are right to conclude that anything less than 19/18/19 would not generate more than 176 (18/17/18 only gives 175) and 19/18/19 needs a good bit more than 8 units high (as you say, 8.6444....)

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?