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Question 7 : What is the minimum number of weights which enable us to weigh any integer number of grams of gold from 1 to 100 on a standard balance with two pans? (Weights can be placed only on the left pan)
To weigh all the numbers we take weights in the ascending powers of 2.
Starting from 20 and 21 and so on up till the necessary weights are present.
The sum of the all the weights up to 2n would be 2n+1 -1
The required maximum weight should be lower than 2n+1 -1.
So for this particular question, if we have to measure weights up to 100kg. We need as many weights in the ascending powers of 2 till the sum is more than 100.
For n = 6, the sum is 63 which is not enough.
For n = 7, the sum is 127, which can be used to measure all the weights.
Hence we need a minimum of 7 weights.
The weights being: 1, 2, 4, 8, 16, 32, 64
Bonus:
If you are allowed to place weights on both the sides of the pan then the powers will be taken in the powers of 3 instead of two. This is because you are also allowed to subtract the weights by placing the weights on the opposite side of the pan. For example when you have the weights 1, 3, 9
You can measure:
1 kg by placing 1 kg on either side of the pan
2 kg by placing a 3kg and a 1kg weight on opposite pans.
3 kg by placing 3 kg on either side of the pan
4 kg by placing 1kg and 3kg on one side of the pan
5 kg by placing 9kg on one side and 1kg and 3kg on the other side
In a similar fashion, we can measure all the weights up to 9+3+1 =13 kgs
The question is " What is the minimum number of weights which enable us to weigh any integer number of grams of gold from 1 to 100 on a standard balance with two pans? (Weights can be placed only on the left pan) "
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