If a shopkeeper can only place the weights on one side of the common balance. For example, if he has weights 1 and 3 then he can measure 1, 3 and 4 only. Now the question is how many minimum weights and names of the weights you will need to measure all weights from 1 to 1000? This is a fairly simple problem and very easy to prove also.
I am the largest in my family of fifty.
The youngest of our family lives separately from the rest of the family and so do I.
I am regarded as harsh and tough.
I am so large, you cut me in two equal halves,
each half would still be larger than the second largest member of the family.
Who am I?