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.
While house hunting in London, I came across a very good leasehold property Discussing the lease the landlady told me:
'The property was originally on a 99 years lease and two-thirds of the time passed is equal to four-fifths of the time to come. Now work it out for yourself and see how many years are to go!
A man had five children. He had $100 with him to give to his children. He decided to start with the youngest child and then give $2 more than each younger child to his next elder child.
For example, if he gives $x to the youngest child, he will give $(x+2) to the next one, $[(x+2) + 2] to the next one and so on.
Can you find out how much did the youngest one receive?