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.
It has five wheels, though often think four, You cannot use it without that one more, You can put things in it, you can strap things on top, You can't find it in the market, but you can still go shop. What is it?
From a pack of 52 cards, I placed 4 cards on the table.
I will give you 4 clues about the cards:
Clue 1: Card on left cannot be greater than the card on the right.
Clue 2: Difference between the 1st card and 3rd card is 8.
Clue 3: There is no card of an ace.
Clue 4: There are no face cards (queen, king, jacks).
Clue 5: Difference between the 2nd card and 4th card is 7.