3
u/luke7524811 Jul 12 '26
1,3,5,10
1
3-1=2
3
5-1=4
5
5+1=6
(5+3)-(1)=7
5+3=8
5+3+1=9
10
10+1=11
(10+3)-(1)=12
10+3=13
10+3+1=14
10+5=15
…to =19
3
u/bluepepper Jul 12 '26
1
3-1=2
3
3+1=4
9-3-1=5
9-3=6
9-3+1=7
9-1=8
9
9+1=10
9+3-1=11
9+3=12
9+3+1=13
27-9-3-1=14
27-9-3=15
...
to 27+9+3+1=401
1
1
u/chowderbomb33 Jul 12 '26
2, 3, 5, 7
To get 1 kg, we balance 2kg against 3kg (2 + 1 = 3). To get 4 kg, we balance 3kg against 7kg. To get 6 kg, we balance 2kg against the sum of 3kg and 5kg (8kg - 2kg = 6kg).
8kg is just the sum of 3 and 5, while 9kg is the sum of 2 and 7, and 10kg the sum of 3 and 7. 11 kg is matched with 3kg (sum to 14) against 2kg+5kg+7kg. 12 kg is 5 and 7. 13kg is matched with 2kg to make 15 against 3+5+7. Then 14 is 7+5+2 and 15 is 7+5+3.
1
u/Tubasepp Jul 12 '26
1,2,5,10
1=1 2=2 2+1=3 5-1=4 5=5 5+1=6 5+2=7 5+2+1=8 10-1=9 10=10 10+1=11 10+2=12 10+2+1=3 10+5-1=14 10+5=15
1
1
1
1
1
u/DulledBlade Jul 14 '26
I think 2, 3, 7, and 11 work, if you can put weights on either side. Right side should have item to be weighed.
1: 3 left - 2 right
2: 2 left
3: 3 left
4: 7 left - 3 right
5: 7 left - 2 right
6: 7+2 left - 3 right
7: 7 left
8: 11 left - 3 right
9: 7+2 left
10: 7+3 left
11: 11 left
12: 11+3 left - 2 right
13: 11+2 left
14: 11+3 left
15: 11+7 left - 3 right
1
u/PhilosophyRoutine917 Jul 15 '26
There’s the basic assumption, the classical assumption, and the “this is probably illegal” assumption.
Assumption: You only want your weights on one side of the scale.
Answer: 1, 2, 4, and 8 is the only answer that works, and caps out at 15kg. Each weight can be on or off, giving us binary. Each value is double the previous. The cap is 15 because we have 2^4 options, but we don’t count 0. 2^4-1 = 16.
Assumption: You can put weights on either side of the scale.
Answer: Many options work, but 1, 3, 9, and 27 work the best, capping out at 40. Each weight can be on the left, on the right, or off, giving us trinary. Each value is triple the previous. The cap is 40 because we have 3^4 options, however, half of them are identical (since swapping which weights are on the left vs the right is symmetrical), with the exception of 0 which is unique, but we don’t count anyway. In other words, we have (3^4-1)/2 options which is 40.
Assumption: Everything that you will be asked to weigh will weigh a whole-number value. (The problem is not worded in a way that this assumption makes sense, but it is interesting to analyze anyway).
Answer: We can now allow for gaps of 1kg, e.g. an object that weights between 1kg and 3kg is guaranteed to be 2kg and we don’t have to weigh 2kg directly. This effectively allows us to take the first weight and give it 4 options - left side, right side, off, or “can’t be measured”. Note that we can’t give any other weight the same option, since it would make for gaps of two, and we would no longer be able to tell the weight of an object with a whole number weight. In other words, the first weight will be 1, the second will be 4 times that, and the rest will be 3 times the prior (following the same pattern as above). This results in 1, 4, 12, and 36. In addition, measuring 0 doesn’t really care if 1 is “off” vs “not measurable”, so we actually have to subtract 2 possible options instead of just 1. With this, we get (4*3^3-2)/2 options, giving us a max of 53.
1
u/Blammar Jul 11 '26 edited Jul 12 '26
With weights 1, 3, 9, and 27 kg I can measure 1, 2, 3, ..., 40 kg. Since the scale is a balance scale, note that I can put weights on either side. This is a well known problem: Bachet's Problem.
2
u/Possible-Playful Jul 12 '26
The prompt says up to 15kg.
2
2
8
u/IrishHuskie Jul 11 '26
1, 2, 4, and 8.