Andrew is 7 and Anne is 13.

EoinkMember

431 replies

EoinkMember

Andrew is 7 and Anne is 13.

stevedMember

Agreed with Eoink. An interesting puzzle in that solving the equations "mathematically" quickly gets complicated, but the solution easily yields when the range of possible solutions is investigated.

Don AtkinsonMember

steved posted:

That’s what I felt. It took a while to realise that the range is limited by “squares” less than 62.

EoinkMember

Don Atkinson posted:steved posted:That’s what I felt. It took a while to realise that the range is limited by “squares” less than 62.

Exactly, it's a very clever question. You look at it for a while trying to find a general solution, decide that brute force is the only practical way, sigh and get ready for a slog, then look at the 62 and realise it's a very manageable set, especially if like me you start at 7 to work down.

EoinkMember

Maybe we should add a stipualtion to challenge Don, the numbers must all be Naim amplifier models (the thought being sparked by the 62 in the one above). So 12, 22, 32, 32.5..., 90, 110, 135...

Actually thinking about it, that'll be pretty limiting, apart from the S1s I think all preamps will end with 2 and most if not all power amps are multiples of 5 with the majority being multiples of 10.

Don AtkinsonMember

**Motorists !!!!!!!!!!!!!**

Myself and Mrs D were walking along a country lane on Tuesday (one of my days off) at a steady 3½ mph when a car overtook us, missing Mrs D by inches !

“Cricky !” said Mrs D, somewhat disturbed. (well, I think she said Cricky !) “He must have been doing 60mph or more !”

“Oh! I doubt it” said I, appreciating that most motorist are very considerate towards pedestrians (it’s a storey – not reality !!!) “We have only taken 27 steps since that car disappeared around that bend ahead. Let’s see how many more steps it takes to reach that bend”

“135”, said I, as we reached the point at which the car had disappeared.

“So what ?” asked Mrs D.

“Well, I can now tell you how fast that car was travelling, assuming it was going at a steady speed”

“And……………?” asked Mrs D

Nick from SuffolkMember

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.

Don AtkinsonMember

**“We have only taken 27 steps since that car disappeared around that bend ahead. Let’s see how many more steps it takes to reach that bend”**

I think I could have worded this part more clearly. From the time the car passed us, to the time it disappeared around the bend, we had taken 27 steps at 3½ mph.

It took us a further 135 steps to reach the bend.

EoinkMember

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

Don AtkinsonMember

**Trains !!!!!!!!!!!!!!!!**

Myself and Mrs D travelled by train from Newbury to Milchester last week and an hour after leaving Newbury something went wrong with the train.

We had to continue the journey at three-fifths of the earlier speed and it made us two hours late at Milchester.

The Conductor mentioned that if the fault had developed 50 miles further on, the train would have arrived 40 minutes sooner.

How far is it from Newbury to Milchester ?

PeakmanMember

Don Atkinson posted:

Trains !!!!!!!!!!!!!!!!Myself and Mrs D travelled by train from Newbury to Milchester last week and an hour after leaving Newbury something went wrong with the train.

We had to continue the journey at three-fifths of the earlier speed and it made us two hours late at Milchester.

The Conductor mentioned that if the fault had developed 50 miles further on, the train would have arrived 40 minutes sooner.

How far is it from Newbury to Milchester ?

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

Don AtkinsonMember

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

Don AtkinsonMember

**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.

Don AtkinsonMember

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 !

Don AtkinsonMember

Nick from Suffolk posted:

Ah Nick,

Let's put the 24.5 mph (vi 21 mph) "near-miss" down to * my* poor wording, rather than

Innocent BystanderMember

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).

Don AtkinsonMember

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 mph** instantly,** then i'm fit enough.....almost as good as winky's Tesla

Ash43Member

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

Innocent BystanderMember

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.

Don AtkinsonMember

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.

Don AtkinsonMember

**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 ?

Don AtkinsonMember

Last edited by Don Atkinson

**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”

HungryhalibutMember

153 and 371.

Innocent BystanderMember

Don Atkinson posted:

Bill & Ben the flowerpot menTwo 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?

Don AtkinsonMember

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

HungryhalibutMember

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

Don AtkinsonMember

Last edited by Don Atkinson

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 ?

Innocent BystanderMember

151

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

Don AtkinsonMember

Hungryhalibut posted:153 and 371.

Neat !!

Don AtkinsonMember

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 ?

Innocent BystanderMember

Last edited by Innocent Bystander

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.

Don AtkinsonMember

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 xxx, which might, or might not fit and might or might not be more than 176........one of these days............

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

Cheers

Don

Don AtkinsonMember

**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.*

Innocent BystanderMember

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 ?

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)

Don AtkinsonMember

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 ?

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

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 BystanderMember

Innocent Bystander posted:Don Atkinson posted:

Bill & Ben the flowerpot menTwo 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?

Don AtkinsonMember

Not yet.

Perhaps others can ?

Don AtkinsonMember

**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 !

Don AtkinsonMember

Don Atkinson posted:

Another Packing ProblemI 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............

Don AtkinsonMember

Innocent Bystander posted:Innocent Bystander posted:Don Atkinson posted:

Bill & Ben the flowerpot menTwo 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*

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