UESP Forums

You have a pair of old-style balancing scales, like the ones in TES:IV.
What is the lowest number of weights you need to weigh every number from one to ten, if each weight is a number, and you only have one of each weight? If clarification is needed just ask.

You need three weights, weighing 2, 3, and 7. Sometimes they are placed on the same side of the scale, sometimes on opposite sides. Here's how you get each number:
1 = 3 - 2
2 = 2
3 = 3
4 = 7 - 3
5 = 7 - 2
6 = 7 + 2 - 3
7 = 7
8 = 7 + 3 - 2
9 = 7 + 2
10 = 7 + 3

I think it's six weights. In addition to those used for 1-10, I multiplied each by 4 (as 40 = 10*4) to get the additional weights of 8, 12, and 28.
1 = 3-2
2 = 2
3 = 3
4 = 7-3
5 = 7-2
6 = 7+2-3
7 = 7
8 = 7+3-2
9 = 7+2
10 = 7+3
11 = 8+3
12 = 12
13 = 8+7-2
14 = 12+2
15 = 8+7
16 = 12+7-3
17 = 12+7-2
18 = 12+8-2
19 = 12+7
20 = 12+8
21 = 28-7
22 = 28-8+2
23 = 28-8+3
24 = 28-7+3
25 = 28-3
26 = 28-2
27 = 28-3+2
28 = 28
29 = 28+3-2
30 = 28+2
31 = 28+3
32 = 28+7-3
33 = 28+7-2
34 = 28+8-2
35 = 28+7
36 = 28+8
37 = 28+12-3
38 = 28+12-2
39 = 28+12-3+2
40 = 28+12

It would be five, since you can already make 8 with the previous answer's weights.
2, 3, 7, 12, 28

The first step would be to figure out how many weights would be required. Those of you who were paying attention have probably already realised that minimum number of weights is four (I told Musicman that it could be solved with less than five, and it certainly can't be done with three), but we gain some extra information from actually doing the calculations, so here they are.

This formula will give you the total number of possibilities:
N = number of weights
X = number of weights used (you need to run through all values of X)
Px = number of possibilities for that X value
(! = factorial eg. 5! = 5*4*3*2*1. I think it's fact(n) in excel)

Px = N! / (N-X)! / X! * 2^(X - 1)

A short explanation of why that formula works

Using that formula, we find that the number of possible combinations for 3 weights is 13 - far too few. With 4 weights, we have exactly 40 combinations, and with 5 weights, 121. This seems to suggest that it's possible with just 4 weights - as long as no two combinations reach the same number, and the highest possible weight is 40 (all weights combined = 40kg).

There are a number of patterns that can be used to ensure that no combinations reach the same number (and some of them do quite well, missing just one or two numbers), but only one of them where the combined weight is 40kg. That combination, which is also the correct answer, is: 1, 3, 9, 27. Here's how each of the weights is reached using these numbers:

a/a-2 + 2/3 = 2/a-2

Ø
Because solving the problem gives us 2. But a cannot equal 2; the a ≠ 2 condition is provided by the problem itself, since we cannot divide by zero.

In one town, there lived two eccentrics, Chuck and Huck. Chuck was completely unable to tell the truth on Mondays, Tuesdays and Wednesdays, though he was invariably truthful throughout the rest of the week. As for Huck, he always lied on Tuesdays, Thursdays and Saturdays but told the truth on all other days.
One day, they met a stranger, who asked one of them,
‘What is your name?’
‘My name is Chuck,’ was the answer.
The stranger asked again,
‘What day is it?’
The same interlocutor replied,
‘Yesterday was Sunday’.
‘And tomorrow will be Friday,’ the second man piped in.
The stranger addressed the next question to the second man,
‘Are you sure you are telling the truth?’
‘I always tell the truth on Wednesdays,’ he answered.
The stranger was then able to determine which one of his interlocutors was Chuck and which was Huck, and also what day it really was.
How did he do it?