Weights and measures
 

:rolleyes: A weight, of a material that can easily be worked upon, be it iron, wood etc. of precisely 81 pounds is given to you.
Assuming you have the requisite tools and capability, can you divide it up into 5 parts such that you could weigh any weight from 1 to 81 Lbs. on a pair of scales exactly ? There is no need of fractions.
There is no catch in this problem and for the Europeans it makes no difference to the problem if the weights are in Kg.
If you get it you may spoilerise your answers. Beware of counter- examples.
Mally :coffee:

[QUOTE=mfgoode]:rolleyes: A weight, of a material that can easily be worked upon, be it iron, wood etc. of precisely 81 pounds is given to you.
Assuming you have the requisite tools and capability, can you divide it up into 5 parts such that you could weigh any weight from 1 to 81 Lbs. on a pair of scales exactly ? There is no need of fractions.
There is no catch in this problem and for the Europeans it makes no difference to the problem if the weights are in Kg.
If you get it you may spoilerise your answers. Beware of counter- examples.
Mally :coffee:[/QUOTE]
[spoiler]Think balanced-ternary notation, in which the digits are -1, 0 and +1.

Given a portion of weight 1, you can form weights of 0 and 1, depending on whether
you put it on a scale pan or not.

Given another portion of weight 3, you can add or subtract either of the
previouslyconstructible weights according to whether you put it on the same pan
as the 3 weight or not. I.e. You can form 0, 1, 2, 3 and 4.

Likewise, a portion of weight 9 can have any of the previous weights
added or subtracted. This give you the range 5 through 13.

A portion of weight 27 now takes you as far as 40 by the same construction.

So far, we have four weights (1, 3, 9 and 27), for a total of 40, leaving 41.
Adding any of the previously constructible weights to this one alllows you
to reach 81, which is the required target.[/spoiler]


Paul

This particular problem was always interesting to me because of the apparent pattern of the first 4 numbers but then the 5th number does not fit the apparent pattern. If I had stated this problem, I would probably have said the original weight to be cut up was exactly 121 pounds.

Extrapolated still further, this pattern always yields an odd number but can it ever yield a prime number other than 3?:yawn:

And carried to another level, if you take each of the "weights" and either add 2 or subtract 2, you will get a very high proportion that one or the other or both is a prime number. Is this an application of the strong law of small numbers?

If you want to play with excel a bit, put the number "1" in cell A1 and then paste this formula in cell A2 "=SUM($A$1:A1)*2+1" and then copy cell A2 and paste down a dozen or two cells.

Fusion

[QUOTE=xilman][spoiler]Think balanced-ternary notation, in which the digits are -1, 0 and +1. [Unquote]
:surprised Good thinking and well put Paul.
The easier question now is with the restriction of only one scale pan what is the least possible number of weights to weigh any weight from say 1- 40 Lbs?
Could these questions be generalised?
That should partly answer Fusion Power's question. :question:
Mally :coffee:

wouldn't a one scale pan force the object being weighed to be on that scale, so where would we put the weights to "weigh" the object?

these types of scales are generally the ones that give a reading all their own without weights, so i'm not sure how useful that would be to us

[QUOTE=mfgoode][QUOTE=xilman][spoiler]Think balanced-ternary notation, in which the digits are -1, 0 and +1. [Unquote]
:surprised Good thinking and well put Paul.
The easier question now is with the restriction of only one scale pan what is the least possible number of weights to weigh any weight from say 1- 40 Lbs?
Could these questions be generalised?
That should partly answer Fusion Power's question. :question:
Mally :coffee:[/QUOTE]

Assuming what you wanted to say, but not what you did say, I'd guess that to weigh 1-40 Lbs we need
[spoiler]6 weights, using only powers of 2. It's the number of bits in the number of max Lbs.[/spoiler]

I got mostly the same answer but with a different reasoning!

[spoiler]
I noticed that since we're dealing with 81=a+b+c+d+e and we're
allowed to use +/- a +/- b +/- c +/- d +/- e to create any
specific weight, then we can set d=3 and e=2 since this allows
for the value one and follows a kind of rule. The 'rule' is based
on the idea that the set {-1,0,+1} has 3 members. 81 needs to
be broken into powers of three for most of it's 'coins' aka parts.
I used {-1,0,+1} because we're dealing with two sides of a
scale - otherwise there is no solution since log(81)/log(2)=~6.3
which is greater then the 5 coins we're limited to.

This leaves a+b+c=81-5=76.
If I use c=9 and b=27 which seems reasonable since we're
dealing with an exponetial scale based on 3, then we have:
a=81-b-c-d-e
a=81-27-9-3-2=76-36=40
[/spoiler]

Here's my table so far of combination to get each sum:
[spoiler]
1=3-2
4=9-3-2
5=3+2
6=9-6
7=40-27-9+3
8=40-27-3-2
10=9+3-1
11=9+2
12=9+3
13=40-27
14=40-27+3-2=9+3+2
15=27-9-3
16=27-9-2=40-27+3
17=40-27+9-3-2
18=27-9
19=27-9+3-2
20=27-9+2
21=27-9+3
22=40-27+9
23=40-27+3+9-2
24=27-3
25=27-2
26=27-3+2
28=27+3-2
29=27+2
30=27+3
31=27+9-3-2
32=27+3+2
33=27+9-3
34=27+9-2
35=27+9-3+2
36=27+9
37=27+9+3-2
38=27+9+2
39=27+9+3
41=40+3-2
42=40+2
43=40+3
44=40+9-3-2
45=40+3+2
46=40+9-3
47=40+9-2
48=40+9-3+2
49=40+9
50=40+9+3-2
51=40+9+2
52=40+9+3
53=40+27-9-3-2
54=40+9+3+2
55=40+27-9-3
56=40+27-9-2
57=40+27-9-3+2
58=40+27-9
59=40+27-9+3-2
60=40+27-9+2
61=40+27-9+3
62=40+27-3-2
63=40+27-9+3+2
64=40+27-3
65=40+27-2
66=40+27-3+2
67=40+27
68=40+27+3-2
69=40+27+2
70=40+27+3
71=40+27+9-3-2
72=40+27+3+2
73=40+27+9-3
74=40+27+9-2
75=40+27+9-3+2
76=40+27+9
77=40+27+9+3-2
78=40+27+9+2
79=40+27+9+3
80=40+27+9+(4)
81=40+27+9+3+2
[/spoiler]

Boy was I mad when I got to 80... I was close, though. The solution was correct for all but one case.

The idea with the *censored* of *censored* values for all but one of the 5 is the only solution?

weights and measures.
 

[QUOTE=fetofs][QUOTE=mfgoode]

Assuming what you wanted to say, but not what you did say, I'd guess that to weigh 1-40 Lbs we need
[spoiler]6 weights, using only powers of 2. It's the number of bits in the number of max Lbs.[/spoiler][/QUOTE]
:smile: Hats off to you fetofs. You are the only person I find, who restricts himself strictly to the problem at hand, and gives the correct answers.
Keep it up Buddy To make it a bit clearer, one has to use 1 (2^0) onwards.
Mally :coffee:

[QUOTE=nibble4bits]I got mostly the same answer but with a different reasoning!

[spoiler]
:sad: In such problems the number 1 is preferable to (2 and 1)
|Any way keep it up. "Go On! and Faith will come to you" d'alembert
Mally :coffee:

What's funny is that if we were talking about a currency then it's perfectly fine. The special case of 80 needs just another "2-cent" piece.

There's a general plan: [spoiler] Find out the base of your log which in this case is 3.
81 is 3^4 so obviously 5 coins is enough for the powers 0 through 4 of 3. [/spoiler]

Try this problem for 100 cents using only apportioned coins (weights) and only using one side of the scale for the weights and the other for the object being weighed.

Weights and measures.
 

[QUOTE=tom11784]wouldn't a one scale pan force the object being weighed to be on that scale, so where would we put the weights to "weigh" the object?

these types of scales are generally the ones that give a reading all their own without weights, so i'm not sure how useful that would be to us[/QUOTE]
:smile: Im sorry Tom that was a bit confusing. The scale used is an ordinary two scale pan but instead of utilising both pans to put the weights on ,one can only use one of the two pans. But both pans have to be used.
I hope it is now clear.
Mally :coffee: