Brain Teaser No 1

Oranges (again)

So, just so as we're all working to the same figures, and so the rest of the forum who occasionally peep in here don't feel left out, let me quote the figures so far.

For a square pyramid we could have

70 layers containing 116,795 oranges
broken down into
69 layers containing 111,895 oranges
and
24 layers containing 4,900 oranges

OR for the ones who don't like so much juice

55 layers containing 56,980 oranges
broken down into
45 layers containing 31,395 oranges
and
42 layers containing 25,585 oranges

These were the only solutions I found with the number of layers less than 100.

Now we are next interested in the height of the stack of 70 oranges (and the subsequent 69 stack)

So see my next post

Cheers

Don

Oranges (yet again!!)

There are lots of factors that we would need to consider before doubting your grocer.

Things like the squashiness of oranges under load for a start. Remember, even steel compressess under load !!

Also, did he form the first layer with each orange JUST touching its neigbours or were they squashed up or did he leave small gaps between each orange.

Assuming we are dealing with hard, non-squashable balls that just touch each other, then of course your grocer is a liar (lets not beat about the bush !!)

70 layers or even 69 layers won't fit between a 3' high table and a 15' high ceiling

See my next post for details

Cheers

Don

Oranges (again and again and again)

So we have a 12' high space. Equals 144 inches.

Ball diameter = 3"

Vertical height of each intermediate layer = 2.12132 inches
(OK I know its a nice number with a 3 and a 2 and a square root sign)

So 70 layers would be 149.37 inches high (yes, I know this is a little bit more than 70 times 2.12132, but it would be, wouldn't it)

And 69 layers would be 147.25 inches high

And 68 layers would be 145.13 inches high

So on balance, I would suggest your grocer might be having you on. Or he is using squashy oranges, by which I guess the bottom ones would be squashed by about 1/8 th inch and the top ones virtually unsquashed

Now all these figures were done long-hand on my knee-board, so they might be a little bit out.

All errors and correction gratefully acknowleged........

But a neat set of brain teasers non the less. thanks Matthew, thanks Lynn

Cheers

Don

Lynn,

Wouldn't 69 levels of 3" oranges actually be 68 x 2.12132" + 3

Yes

I think you will find the answer to that sum is 147.25"

If you had assumed the height was 69 x 2.12132" I think you would have got 146.37"

I think you will find my calcs take account of the top and bottom oranges being "vertical"

But, as I say, my figures are hand calcs and my brain was only intermittently working this problem so I might have screwed up somewhere, but I hope not.....

Neat problem, lots of little "twists" to trap the unwarey. Now I think your grocer must have long arms or he dangles from the ceiling to place the top layers, or perhaps uses a set of ladders propped to the rafters.....(oh no, not Bam's ladder problem again !!!!)

Cheers

Don

more oranges?

Don, Lynn and Matthew,

This really isn't a teaser, since I do not have an answer, but I was wondering if it possible to restack a pyramid (of base p) into a tetrahedron (of base t)? Mathematically, are there solutions to

t(t+1)(t+2) = p(p+1)(2p+1)

Obviously there are no whole number solutions for p=t. In fact I think I can show that there are no positive whole number solutions for t=p+1 or t=2p+1. But are there other viable solutions? Feel free not to think about this if you are now dreaming of orange pyramids,

cheers

Dan

Lynn,

Sorry, I missed a step - there's a factor of 1/6 on both sides of the equation that I canelled. To clarify (if I'm not wrong):

#oranges in a tetrahedron of base t is
t(t+1)(t+2)/6

#oranges in a pyramid of base p is
p(p+1)(2p+1)/6

These total agree with your calculations.

I cant seem to get your formula that implies you can take a pyramid of base n and get 2 tetrahedrons of base n and n-1. Can you clarify?

cheers

Dan

n(n+1)(2n+1) - n(n+1)(n+2) = n(n+1)(n-1)
Neat!

quote:

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If we have this settled, should be a simple step to answer my grocer's query!

---

Your grocer needs to sell (n-1)n(n+1)/6 oranges, I think.

Well after a little bit of coding, I havent been able to find a number where you can restack a pyramid into a tetrahedron for levels up to 1 million. Anyone care to prove it can't be done at all?

cheers

Dan

One!

quote:

---

Originally posted by Matthew T:
One!

---

OK, I'll give you that one if you can show me
an orange in the shape of a tetrahedron (or pyramid). Perhaps the oranges on bottom of the grocers stack look this way!

cheers,

Dan

A few loose ends to sort out

Browsing back through the last couple of pages, I realised there are a few loose ends to sort out.

Law of Averages
300SET
Explaining the height of a stack of oranges
Proving that a square pyramid can't be turned into a tetrahedron

There might be more

I can sort out the first and third but am struggling with the last one

I think we need a lawyer for the second one (any offers ?)

cheers

Don

300SET

Having just consulted my lawyer, she points out that the replacement tubes have a higher serial number than the origonals (well, the original boxes to be more precise)

Possibly the store keeper claimed the origonals had failed and he had 'replaced' them with newer, and by implication, better tubes. Or perhaps a careless customer or cleaner had broken the originals etc etc.

This, m'lord, end the case for the defence.........

Cheers

Don

The Pyramids

Or

The thickness of a layer of oranges

This answer is going to be in four bits because I have'nt figured out how to reduce my diagrams to small KBs

cheers

Don

This is the boring bit, ie the words
Five oranges, centres A, B, C, D, E

Orange e is not shown for clarity but sits in the natural hollow formed by oranges a,b,c,d. See diagram in next post.

We need to calculate the height of the pyramid, that is from the mid-point of FG to the apex A.

Triangle ABC is equilateral, side 3"
By Pythagoras, the height AF of triangle ABC, which is also the slope length of the face of the pyramid is 3*(√3)/2

NB it's not necessary to find AF. Finding (AF)2 [which = 27/4] is enough

Triangle AFG is isosceles, side 3*(√3)/2; base 3
By Pythagoras, the height of triangle AFG, which is also the height of the pyramid is 3/√2

cheers

Don

Next diagram

Cheers

Don

And the last diagram,

Cheers

Don

quote:

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Originally posted by Don Atkinson:
Proving that a square pyramid can't be turned into a tetrahedron ... I can sort out the first and third but am struggling with the last one

---

Don, you are not the only one - it seems to be a simple problem but for the life of me I can't see a proof.

cheers

Dan

Ooops!!

I spotted an error in my text about oranges above.

The above text reads : Orange e is not shown for clarity but sits in the natural hollow formed by oranges a,b,c,d. See diagram in next post.

It SHOULD have read :Orange a is not shown for clarity but sits in the natural hollow formed by oranges b,c,d,e. See diagram in above post.

Sorry for any confusion

Cheers

Don

My solution to the Law of Averages

Cheers

Don

Dan, Don

I think the issue with the proof is that there may well not be one. It is very difficult to find a proof when the numbers are constrined to being integers, therefore using your equations will most likely prove hopeless.

Will continue to think on it for a while.

Matthew

An observation...
Lynn has pointed out that a pyramid can be restacked into two tetrahedra - one of the same height as the pyramid and another one with one less level. Therefore the problem is equivalent to solving ft(m) = ft(n) + ft(n-1) (m,n positive integers). Since ft is a cubic, I am reminded of Fermat's last theorem - there exists no solution to a^n = b^n + c^n for n>2 (IIRC). Hopefully solving this problem isn't of similar difficulty! If it is we could be here a while!

cheers,

Dan

Orange pyramids. Some thoughts.

The conjecture is that none of the infinite series of square pyramids of oranges can be turned into a tetrahedron pyramid.

The number of oranges (n) in each, based in the number of layers, is

Tetrahedron n = t(t+1)(t+2)/6 where t=number of layers
Square n = p(p+1)(2p+1)/6 where p=number of layers

In the tetrahedron we have three consecutive numbers as factors of n,
hence one (but only one) of the factors must be divisible by 3. Also, at least one (but possibly two) of the factors must be even and divisible by 2. Hence t(t+1)(t+2) is definitely divisible by 6. Which we knew all along.

In the square pyramid we have two consecutive numbers as factors of n, so one (and only one) of the factors is even and divisible by 2. The factor (2p+1) is always odd, and increases as the series 1,3,5,7.....However, I can't see how to demonstrate that at least one of the three factors is divisible by 3, which must be the case if p(p+1)(2p+1) is divisible by 6 (which we seem to think is the case).

As I said, some thoughts!!!!

Cheers

Don

A quick diversion

I heard this one on the radio this weekend - it is from the Car Talk radio show on NPR. I thought we might need an easy one to keep us going.

quote:

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A family of four and their dog are trapped on an island, when rising floodwaters tear out the bridge that they had used just a few hours earlier. When they had just given up hope the son says, "I've got a small boat and oars."

But their joy was short-lived. The manufacturer's instructions printed on the boat stern tell that the boat can carry only 180 pounds.

The dog is the only one who can swim.

The father weighs 170. The mother says she weighs 130 but it's more like 155. The son is 90 pounds, the daughter is 80, and the dog is 15.

So, what's the fewest number of crossings they have to make to save everybody?

---

cheers

Dan

Assuming that saving them involves getting off the island it could be done in one crossing assuming that they don't mind getting wet. If they don't want to get wet then they better hope the dog has enough sense to tug the boat back to the island else they are in trouble!

Matthew

I have a box 5cm x 5cm x 10cm and spherical balls of diameter 1cm. What is the maximium number of balls can I fit in the box?

Matthew

350