unforum

TopGun2 · Boot Joined: 30 Nov 2005 Posts: 30 Location: Cambs, UK

Have spotted the 23 now, d'oh!

Will play with matrices later.

**Posted:** Sun Apr 09, 2006 12:28 pm

_________________
Audere Est Facere

Phill4269 · Greenhorn Joined: 24 Feb 2006 Posts: 7 Location: London

The minimum right had side moement is 98 (2x22 + 6x9). This means that the minimum moment divisible by 5 is 100. This can be achieved uniquely on the right hand side.
On the left hand side I have found at least 3 solutions which give a moment of 20x5=100.
Two have been incorrect the third I will try tomorrow. The problem is that all solutions must add up to 20Kg - so can there be only one correct solution?

If there is an error in the card, there isn't much scope! Maybe the answer processing is wrong.

I have tried getting a hint but got no response. Has anyone managed to get a hint?

**Posted:** Sun Apr 09, 2006 12:48 pm

gagravarr · Greenhorn Joined: 09 Apr 2006 Posts: 4

**Posted:** Sun Apr 09, 2006 1:01 pm

hot flan · Guest

as far as i can figure, it's impossible Confused

using magma's excel file the lowest total weight possible is 52kg, and there is definitely more than one solution for this. unless some of the weights can be 0 I can't see how this can work

**Posted:** Sun Apr 09, 2006 2:12 pm

Magma · Veteran Joined: 10 Jan 2006 Posts: 119

Well if 0 is possible, then the minimum configuration of the system is 0 for each weight. But that would kinda defeat the whole concept of "Weights" wouldn't it? Smile

**Posted:** Sun Apr 09, 2006 3:25 pm

Stratman

Getting time to call this a day and head for the pub.
100 each side can be done and i dont see how it could be less.

Hope this doesn't sound dumb, but going back to the way Friedman's puzzles usually work...(Sorry, Austin, had another thought on this)...
http://www.stetson.edu/~efriedma/weight/
Thinking about this, using the weights 1,2,3,4...25 could be considered to be the lightest combination as it is worded. Using any combination of weights could mean the lowest weight would not necessarily have to be 1 and they would not have to be consecutive - so higher combinations would be possible.
If so, it could alternately mean 1,2,3,4,5 for the first weight tree and 1,2,3,4...20 for the second.
But of course the card should make it clear that no two weights can be the same value - and it doesn't. It would not be the first card with a mistake on it.
It is the only thing I can think to explain why our 100/100 balances dont get a solve. It also makes the puzzle much harder and probably more worthy of a purple.

**Posted:** Sun Apr 09, 2006 4:33 pm

_________________
There Ain't Half Been Some Clever Bastards...Ian Dury and the Blockheads (1978)

a11420151425 · Boot Joined: 26 Nov 2005 Posts: 10

program

**Posted:** Sun Apr 09, 2006 5:23 pm

Preon neKaskth · Boot Joined: 08 Apr 2006 Posts: 16

I think by "lightest possible combination", the card refers to the mass of the weights: the lightest of which i've completed the big tree with is 55.

It ends up with the numbered weights being:

4. 2
5. 2
6. 2
7. 1
8. 3
9. 3
10. 6
11. 2
12. 1

I can't see anyother way to do it. But a nice puzzle, I like it.

**Posted:** Sun Apr 09, 2006 6:37 pm

Grizy

Preon please read previous threads.
There are at least 2 ways of doing it with 52Kgs.
But we still haven't solved it. Sad

**Posted:** Sun Apr 09, 2006 6:43 pm

Preon neKaskth · Boot Joined: 08 Apr 2006 Posts: 16

Aha... whoops... didn't register as I went through it... never mind.

**Posted:** Sun Apr 09, 2006 6:47 pm

a11420151425 · Boot Joined: 26 Nov 2005 Posts: 10

help

**Posted:** Mon Apr 10, 2006 8:30 am

Stratman

six

**Posted:** Mon Apr 10, 2006 8:35 am

_________________
There Ain't Half Been Some Clever Bastards...Ian Dury and the Blockheads (1978)

a11420151425 · Boot Joined: 26 Nov 2005 Posts: 10

thanks

**Posted:** Mon Apr 10, 2006 8:36 am

gagravarr · Greenhorn Joined: 09 Apr 2006 Posts: 4

Card has a problem

**Posted:** Mon Apr 10, 2006 8:44 am