Brain Teaser No 1

Posted by: Don Atkinson on 16 November 2001

THE EXPLORER

An explorer set off on a journey. He walked a mile south, a mile east and a mile north. At this point he was back at his start. Where on earth was his starting point? OK, other than the North Pole, which is pretty obvious, where else could he have started this journey?

Cheers

Don

Posted on: 02 December 2003 by Don Atkinson
Minky,

you'll probably find Mick Parry and one or two others still posting, either here or on Mana....

Cheers

Don
Posted on: 02 December 2003 by Don Atkinson
Chess? or draughts?

How many squares are there on a standard chess board?

I don't mean just the obvious single squares that are coloured either black or white, but these and all the 2x2 and 3x3 etc combinations that can be formed by combining the basic squares.

Cheers

Don
Posted on: 02 December 2003 by Dan M
204

cheers

Dan
Posted on: 03 December 2003 by Matthew T
I agree with Dan.

Sum of squares and all that.

Matthew
Posted on: 04 December 2003 by Don Atkinson
Dan, Matthew,

Well done!!

204 is good.

"Sum of squares (and all that)" is clearly correct and shows we all share a common awareness of the solution.

So for the benefit of other readers, would one of you like the honour of explaining how we get 204 and the general princple for any size 'chessboard', or should I?

Cheers

Don
Posted on: 04 December 2003 by Two-Sheds
for a chessboard (8x8), then the number of single squares (1x1) on the board is obviously 8*8 or 8^2 which is 64.

If we then look at how many squares are on the board that are 2x2. If you take the top 2 rows of the board then we can easily see we can fit a 2x2 square in 8x2 section 7 times. Likewise going down the board again we can see we can get a 2x2 square in a 2x8 section 7 times as well. So for the whole board we can get 7*7 or 7^2 2x2 squares into an 8x8 board.

Here we can start to see the pattern emerge. The total number for an 8x8 board will be:
8^2+7^2+6^2+5^2+4^2+3^2+2^2+1^2 = 204

likewise for a board nxn the formula would be:

n^2+(n-1)^2+...+1^2
Posted on: 05 December 2003 by Dan M
2-sheds,

Nice description. I'd like to add that another way to write

n^2+(n-1)^2+...+1^2

is n(n+1)(2n+1)/6

cheers,

Dan
Posted on: 05 December 2003 by Don Atkinson
another way to write

n^2+(n-1)^2+...+1^2

is n(n+1)(2n+1)/6

Nice one Dan.

Does this ring any bells here??

Like way back when we were talking about orange sellers and their displays......

Cheers

Don
Posted on: 05 December 2003 by Dan M
Don,

Oh yes, I remember those fruit pyramids and tetrahedrons well!

cheers

Dan
Posted on: 07 December 2003 by Don Atkinson
Confession week !

So, those of you read the main hifi page on this forum will probably be aware that me and Mick Parry are hifi frauds because we didn't realise one of Mick's speaker cables had be wired up out of phase.......

Well! I have another confession to make this week.

It's to do with odd numbers and perfect squares. You see, I have just realised that if you......oh!....but no doubt the relationship is so obvious; and was explained on day 3 at primary school (that's the day I missed because of a cold !); that you will all be falling over yourselves to tell me all about this relationship between the odd numbers (1, 3, 5, 7....) and the square numbers (1, 2, 4, 9, 16....)

Cheers

Don
Posted on: 07 December 2003 by John Channing
tell me all about this relationship between the odd numbers (1, 3, 5, 7....) and the square numbers (1, 2, 4, 9, 16....)

Well if you sum the odd numbers you get the squares:
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
John
Posted on: 07 December 2003 by Don Atkinson
Well if you sum the odd numbers you get the squares:

and

the first odd number = 1^2

the first two odd numbers = 2^2

the first three odd numbers = 3^2

the first four odd numbers = 4^2

etc

Cheers

Don
Posted on: 07 December 2003 by John Channing
The sum of the first n terms of an arithmetic progression is

S = n/2(2a + (n - 1)d)

where a is the first term and d is the difference between the terms. In the case of the odd numbers, a=1 and d=2 giving the observed result

S = n^2
John
Posted on: 07 December 2003 by Don Atkinson
The sum of the first n terms of an arithmetic progression is

neat one John.

Myself and, I'm sure, many others here, are familiar with this expresion and with a bit of a push could re-derive it again........ok, perhaps a big push.....

But the impersonailty of this, and many other expressions, prevents us (ok perhaps just me) from seeing simple relationships like the odd numbers and the square numbers.....

......just a few rambling thoughts that I thought you might (not) be interested in....

Cheers

Don
Posted on: 07 December 2003 by John Channing
Myself and, I'm sure, many others here, are familiar with this expresion and with a bit of a push could re-derive it again........ok, perhaps a big push.....

I'm fairly sure when I did A-level maths I could derive this formula and the one for a geometric progression, but I had to look it up this time so I pulled out my copy of Mathematical Methods in the Physical Sciences by Mary L. Boas (which contains pretty much everything about Maths you need to know even up to PhD level in Physics or Engineering) and to my surprise it didn't contain it! So I had to resort to the good old internet which came up with the result straigh away.
John
Posted on: 09 December 2003 by Don Atkinson
For chocaholics....

The picture shows a bar of chocolate. (really). There are 24 pieces in the block. (6 x 4)

I intend to snap the bar into 24 individual pieces in the usual way, ie by snapping it along the 'valleys'

What is the minimum number of times will I need to snap, to create the 24 individual pieces.

Cheers

Don

PS Our American cousins can pretend ita a waffle, if that helps.....
Posted on: 09 December 2003 by Minky
5 ?
Posted on: 09 December 2003 by Dan M
quote:
Originally posted by Minky:
5 ?

Is stacking allowed?
Posted on: 09 December 2003 by Don Atkinson
Minky,

I should perhaps add, that once a piece, or a strip, has been snapped off, it, and any other pieces or strips have to be broken down individually eg you can't hold two strips together and snap them simultaneously

cheers

Don
Posted on: 09 December 2003 by Minky
Oki Doki, then I volunteer to be the fall guy, because I can't see any way to do this with less than 23 straight snaps. If it's possible to snap off a piece in the middle of the edge of our cyber-bar (in the real chocolate world this would be tricky and messy) you could take out three non-adjacent peices from the long edge and snap the other three in one go, repeat twice more then break up the last length. This is 17 snaps.
Posted on: 09 December 2003 by Dan M
I'm with minky on this one -- any way I sliced (or snapped) it, I got 23.

cheers

Dan
Posted on: 11 December 2003 by Matthew T
I can't see that it is possible to do this with any other then 23 breaks.

Matthew
Posted on: 12 December 2003 by Don Atkinson
Minky, Matthew and Dan,

Ok. 23 is the answer to the MINIMUM number of breaks.

It is also the answer to the MAXIMUM number of breaks (although i didn't get round to asking that bit !)

So, as Matthew pointed out, it isn't possible to do it with anything other than 23 breaks.

Well done guys,

Cheers

Don
Posted on: 12 December 2003 by Don Atkinson
Continuing with the waffle

or choc bar.....

The general formula for breaking a bar of chocloate into pieces is....?

Cheers

Don
Posted on: 14 December 2003 by Don Atkinson
Nice answer, Omer.

Cheers

Don