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.
Two friends were stuck in a cottage. They had nothing to do and thus they started playing cards. Suddenly the power went off and Friend 1 inverted the position of 15 cards in the normal deck of 52 cards and shuffled it. Now he asked Friend 2 to divide the cards into two piles (need not be equal) with equal number of cards facing up. The room was quite dark and Friend 2 could not see the cards. He thinks for a while and then divides the cards in two piles.
On checking, the count of cards facing up is same in both the piles. How could Friend 2 have done it ?