Weights on a Balance

Logic Level 3

Suppose you have a balance and an unknown weight of m kg m \text{ kg} , such that 100 m > 0 100 ≥ m > 0 . If you are allowed to craft a weight of any known integral mass, what is the least amount of weights needed to be crafted to determine the mass of the unknown weight, given that you are allowed to put the weights on both pans of the balance?

Note: m m is a positive integer


The answer is 5.

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4 solutions

Abin Das
Jan 7, 2020

The unknown weight is of the range 1-100 kg. Hence to determine the unknown weight, we will need a set of weights that can determine all the weights from 1 kg to 100kg (or 1kg to 99kg. If none of them matches, it implies that m is 100kg. Correct me if I am wrong).

Consider one of the weights as 67kg (I will explain later why I chose this weight). Assume we already have some set of weights that can be used to measure all the weights from 1kg to 33kg. Hence we can measure all the weights in the range (67+1)kg to (67+33) kg and (67-33)kg to (67-1)kg i.e, 68kg-100kg and 34kg-66kg. We already have 67 kg weight and a set of weights that measures from 1 kg-33kg. Thus we can measure all the weights from 1kg-100kg.

The next step is to find the set of weights for (1-33)kg. This can be done in the similar way as above. In the above scenario, I took two-third of 100, i.e, rounded off to 67. Hence the next weight can be 33 * ( 2 3 \frac{2}{3} ) = 22kg . Assuming that we already have a set of weights that can measure (1-11)kg, the above mentioned scenario will follow.

Next step is for (1-11)kg . 11 * ( 2 3 \frac{2}{3} ) is rounded off to 7kg . Remaining (1-4)kg.

4 * ( 2 3 \frac{2}{3} ) rounded off to 3kg . Remaining 1kg .

Hence the set of weights that can exactly measure all the weights from 1kg to 100kg is 1kg, 3kg, 7kg, 22kg, 67kg. i.e, 5 weights

Hence the answer is 5 \boxed{5} .

Eg : Assume m = 85 kg. Put m, 1kg, 3kg on one pan and 67 kg,22kg on the other.

Gary Miller
Aug 28, 2020

As others have said the weights are 1, 3, 9, 27, and 81 or 3^{0}, 3^{1}, 3^{2}, 3^{3}, and 3^{4}

Luca Ng
Aug 21, 2020

1,3,9,27,81 is sufficient For example you would want to weigh 64 kg, you could do 81, 9, 1 on a side and 27 on the other.

Douglas Foster
Dec 17, 2019

The five weights are 1, 3, 9, 27, 81. I did not prove that this was the fewest number of weight, but i could not see how to go lower.

The reason this work is because each weight is the sum of all previous weights, times 2, plus 1. Stopping when you can sum them to 100. In fact these fives weights could determine anything up to 121kg

No set of 4 4 weights can measure 100 100 distinct weights for the following reason. For any weight, there are precisely three possibilities- either it is put on the left pan, or the right pan, or not put on either of the pans. Hence, for any set of 4 4 weights, we can measure at most 3 4 = 81 3^4=81 distinct weights.

Note: The actual number of distinct measurable weights will be less than 81 81 as switching the weights from the left pan to the right pan, and vice versa, does not change the measurable weight.

Abhishek Sinha - 1 year, 5 months ago

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We don't need to be able to measure 100 weights to determine the unknown weight. We only need to be able to measure 50 weights. For example, if we could measure 2, 4, 6, 8, 10, ... we could figure out the unknown weight. If it is heavier than 8 but ligher than 10, we would know it is 9.

Here is the correct explanation why 4 weights are not possible: We can only measure 3 4 = 81 3^4 = 81 different weights. If we can measure m m , we can also measure m -m by switching the pan of all weights. Hence we can only measure 40 positiv weights which is less than the 50 positive weights that are required.

Finnley Paolella - 11 months, 4 weeks ago

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