Brain Teaser No 1

The six sector problem has multiple solutions!
These all work:
1,2,5,4,6,13
1,2,7,4,12,5
1,3,2,7,8,10
1,3,6,2,5,14
1,10,8,7,2,3
1,14,4,2,3,7
John

Whereas the 7 sector problem does not appear to have a solution equal to 43...
John

THE PENTAGON
Don,
If Marcus stands near to one of the points on the pentagon, he can touch 2 walls without having to travel at all. He needs to find a location on an imaginary line to the orange spot, where he can just touch both walls. This location "X" will depend on how long his arms are, but will be about 1 metre from the point of the pentagon, and therefore around 16m from the orange spot. His region will be defined by the perpendicular line from "X" to each of the walls, plus the walls themselves - ie 4 sides. .
Steve D

[This message was edited by steved on MONDAY 07 July 2003 at 14:55.]

quote:

---

Whereas the 7 sector problem does not appear to have a solution equal to 43...

---

The joy of brute force?

It would be nice to do some elegant solving of this, but brutism is what I'm reduced to.

Paul

Here are (the) 8 unique, 8 sector solutions.
1,2,10,19,4,7,9,5
1,3,5,11,2,12,17,6
1,3,8,2,16,7,15,5
1,4,2,10,18,3,11,8
1,4,22,7,3,6,2,12
1,6,12,4,21,3,2,8
1,6,17,12,2,11,5,3
1,8,2,3,21,4,12,6

The joy of brute force?

Yes, well, digital computational force. Over the weekend, I figured out a reasonably well optimised programmatic solution which has thus far yielded quick solutions to all the problems with fewer than 6 sectors, no solution to the 7 solution sector and a handful of soultions to the 8 sector problem.

It would be nice to do some elegant solving of this, but brutism is what I'm reduced to.

I agree. However, my only observations that reduce the complexity of the problem thus far have been:
1) No number in the solution is ever repeated
2) The maximum single number is always less than or equal to (max+1)/2 and it is (Max-1)/2 in some cases.
John

The best I can do with seven sectors...
39=1,2,5,4,4,6,17
John

THE PENTAGON,
Don,
After giving the matter some further thought, I don't think my answer earlier is correct. However, I'll leave it there until I, or anyone else, thinks of anything better!!
STEVE D

How many sides does Marcus' region have ?

10

What is the shortest distance from the orange spot to Marcus' chalk line. ?

10M
John

[This message was edited by John Channing on MONDAY 07 July 2003 at 19:12.]

John,

Last time I looked, I could have sworn I saw a post suggesting six sides to Marcus' chalk line ??

But your latest posting is a big improvement.

Cheers

Don

SteveD,

Marcus' chalk line.

Usual rules apply viz,

If you stand 1m away from each of two walls near a corner, you have to travel 2m (1m there and 1m back) to get to a wall and back. Then another 2m to get to the other wall and back. You then still have three more walls to touch.....

If you stand IN the corner, of course you can touch two walls without travelling....but you still have three more walls to touch.....

No outstretched hands allowed!!!!!!!

Cheers

Don

SteveD (and John)

Of course John has cracked Marcus' chalk line.

If Marcus started from the orange spot, he would travel 10 times the perpendicular distance to a side in order to complete the task.

If he started (say) 5m away from the orange spot, he would also travel exactly 10 times the perpendicular distance to a side in order to complete the task......

Dito if he starts anywhere inside the chalk line.

Outside the chalk line, well thats a different story and a bit further to travel than from inside the chalk line....

Hope this helps

Cheers

Don

Last time I looked, I could have sworn I saw a post suggesting six sides to Marcus' chalk line ??

I quickly realised that the solution lay in perpendicular lines from each wall to the spot. The shape is therefore generated by drawing a perpendicular line at the end of each side in towards the centre. A rough sketch by hand made me think the answer was 6. When I drew it with a ruler I got the correct answer 10 and since no one had attempted the question since my first effort I just amended it. I will attempt a drawing.
John

JohnC,

since no one had attempted the question since my first effort I just amended it

thank goodness....I though I was going up the wall....

I will attempt a drawing. always a good idea...

Cheers

Don

Here is my solution
John

Neat.....especially the orange spot....

Cheers

Don

SteveD,

After giving the matter some further thought, I don't think my answer earlier is correct

a courageous statement Steve.....we're with you on this one....

but it was a good try, and made me realise I have to draft the questions more carefully..

Cheers

Don

Dicey research.

The Naim research department has a very useful piece of kit to help make decisions about which components to use and which type of sound sounds best etc. It's a box, full of standard dice.

(ok paul, graham, Richard and co, I'm only joking....i know you've got a two-headed half crown in there as well)

So tonight, I'm doodling with dice. I have built three separate columns with equal-sized dice. Each dice is placed squarely on the one below. I have read the top of each column as a digit and then read the digits together to form one long number.

For example, if the tops had shown left to right 'one spot'; 'one spot'; 'three spots'....then I would have read 113.

Surprisingly, I found that the number I actually read, equalled the total number of visible spots all around and on top of the three columns of dice.

Equally surprisingly, when I multiplied together the three numbers on the top of each column, it equalled the number of dice I had used to build the three columns.

Spooky....

but what is the total number of spots visible?

Cheers

Don

9 dice, 133. The numbers around the edge of the dice will add up to 14, you therefore have 9*14+1+1+3=133 and 1*3*3=9.
John

John,

9*14+1+1+3=133

I think you have a little typo.......the arithmetic doesn't add up!!!!

But you've got the right answer. Neat

Perhaps its worth reminding people that the opposite faces of a dice add up to 7 ??

Cheers

Don

Perhaps its worth reminding people that the opposite faces of a dice add up to 7 ??

If you didn't know that you'd be stuffed. It was actually quite an easy problem. I wrote down
xyz=3n and
42n + x + y + z = [x][y][z]
where n is the height of the column and the first solution that yields a three digit number is n=3, which was the correct answer.
John

If you didn't know that you'd be stuffed

well, my wife didn't, so she was..........

It was actually quite an easy problem.

yup, I agree...

and we don't actually need to have 3 dice in each column

but.... for a slightly more challenging teaser, (while we give Omer's puzzle some thought), how about finding a similar situation with more than 3 columns of dice......I believe there is only one possibility !!

Cheers

Don

quote:

---

Prove that there is a match with no crossings of lines.

---

Well, I think I can prove this for n=1 Is there a way to show there's a match with no crossings for n+1, given a match with no crossings for n?

cheers,

Dan

Omer,

Thanks for the encouragement, but it's not clear to me that this will work. I can picture the addition of two points requiring a re-matching of all points. Hummm.... Anyone else with an insight on this?

cheers

Dan

There must always be a pair of dots, one black and one white, on the edge of the cluster such that passing a line through them puts all the other dots on one side of the line. Join the dots and then remove them from the puzzle. Eventually you have no dots left. QED?

Or perhaps not, some argument along those lines perhaps?

Paul

but.... for a slightly more challenging teaser, (while we give Omer's puzzle some thought), how about finding a similar situation with more than 3 columns of dice......I believe there is only one possibility !!

With 4 columns, 180 dice and 2536 works.
John