You have 4 weights and with those you have to be able to equal all weights from 1 to 40? Please Help?

Question

4 weights (from 1 to 40), when used together, equal all weights between 1 and 40. Thank you in advance.

4 weights (from 1 to 40), when used together, equal all weights between 1 and 40. (Ex. 17, 20, 23, and 39, it give you 40(17 +23), and 1(17+23-39), etc.. (This example doesn't work out). Thank you in advance.

Answer 1

With 4 weights, there are only 16 different combinations (2^4 = 16) of the four weights being used or not-used. There aren't 40 combinations possible.

Assuming one can either use or not-use a weight, the biggest range of 4 weights would be to use powers of 2: 1, 2, 4, 8. Then you could do every weight from 0 to 15, but that's all.

On a balance scale, it is possible, because the weights can be added OR subtracted OR not used at all. Since placing weights on opposite sides of a balance scale effectively subtracts them, that gives you one more possible usage for each weight, and a bigger number of combinations ((3^n)/2 combinations instead of 2^n). I'm assuming that is what is meant, though it would have been helpful to state that in the question.

In that case, you would make your weights each a power of 3: 1, 3, 9, and 27.

1 = 1 on the left-hand side
2 = 3 on the left-hand side, -1 on the right
3 = 3 on the left-hand side
4 = 1 + 3 on the left-hand side
5 = 9 on the left-hand side, -3 -1 on the right
6 = 9 on the left-hand side, -3 on the right
7 = 9 + 1 on the left-hand side, -3 on the right
8 = 9 on the left-hand side, -1 on the right
9 = 9 on the left-hand side
10 = 9 + 1 on the left-hand side
11 = 9 + 3 on the left-hand side, -1 on the right
12 = 9 + 3 on the left-hand side
13 = 9 + 3 + 1 on the left-hand side
14 = 27 on the left-hand side, -9 -3 -1 on the right
15 = 27 on the left-hand side, -9 -3 on the right
16 = 27 + 1 on the left-hand side, -9 -3 on the right
17 = 27 on the left-hand side, -9 -1 on the right
18 = 27 on the left-hand side, -9 on the right
19 = 27 + 1 on the left-hand side, -9 on the right
20 = 27 + 3 on the left-hand side, -9 -1 on the right
21 = 27 + 3 on the left-hand side, -9 on the right
22 = 27 + 3 + 1 on the left-hand side, -9 on the right
23 = 27 on the left-hand side, -3 -1 on the right
24 = 27 on the left-hand side, -3 on the right
25 = 27 + 1 on the left-hand side, -3 on the right
26 = 27 on the left-hand side, -1 on the right
27 = 27 on the left-hand side
28 = 27 + 1 on the left-hand side
29 = 27 + 3 on the left-hand side, -1 on the right
30 = 27 + 3 on the left-hand side
31 = 27 + 1 + 3 on the left-hand side
32 = 27 + 9 on the left-hand side, -3 -1 on the right
33 = 27 + 9 on the left-hand side, -3 on the right
34 = 27 + 9 + 1 on the left-hand side, -3 on the right
35 = 27 + 9 on the left-hand side, -1 on the right
36 = 27 + 9 on the left-hand side
37 = 27 + 9 + 1 on the left-hand side
38 = 27 + 9 + 3 on the left-hand side, -1 on the right
39 = 27 + 9 + 3 on the left-hand side
40 = 27 + 9 + 3 + 1 on the left-hand side

Answer 2

I believe this is impossible, even if you're only looking for integer weights. Consider the following.

If I start with a weight of 1, that allows me 1 weight, add in a 2nd weight of 2, that would give me 1, 2, or 3. If I then add a 3rd weight of 4, I could get 1, 2, 3 (1+2), 4, 5 (1+4), 6 (2 + 4), and 7 (1 + 2 + 4). I then add a fourth weight of 8, which adds the following possibilities 8, 9 (1 + 8), 10 (2+8), ... 14 (2 + 4 + 8), and 15 (1 + 2 + 4 + 8).

A fifth weight of 16, would allow me to get up to 31, but it would take a 6th weight.

If we're allowed to subtract weights to get to the number we want, this would change things.

Consider we want to weigh an item X which is between 1 and 40 units. If we put only the item on 1 side with the different combinations of weights on the other side, we can only find 15 different weights. But, we could also put weights on the same side as the object, so that would in fact help us calculate X's weight. (a 40 unit weight alone on one side with a 30 unit weight in addition to the item on the other side would let us calculate the item weighs 10 units).

Working on the weights under these circumstances...

See McFate's Answer. 1, 3, 9, 27 is the answer.

Answer 3

1,3,9,27 answer
22=27-(9-1-3). etc.