These problems was posed to me by friends at NBHM Nurture Programme 2007 at IMSc Chennai

Problem 1:

A shopkeeper can only place the weights in one side of the common balance. For example if shopkeeper has weights 1 and 3 then he can measure 1,3 and 4 only. How many minimum weights and name the weights you will need to measure all weights from 1 to 1000.

Solution 1:

Highlight the part between the * symbols for the answer.

*The numbers 1,2,4,8,16... This is optimal as each number has exactly one binary representation. So for making 1000 kg we need up to 1, 2, 4, 8, 16, 32, 64, 128, 512. :) *

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Problem 2:

Same as the first problem with the condition of placing weights on only side of the common balance being removed. You can place weights on both side and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can…