A man has eighty-one cows ( numbered 1,2,3...81 as such). The beauty is that cow no. 1 gives 1ltr of milk, cow no. 2 gives 2ltrs of milk and so on. The man wants to equally distribute the cows among his nine sons so that each one of them gets the same quantity of milk.
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.
The day before the 1996 U.S. presidential election, the NYT Crossword contained the clue “Lead story in tomorrow’s newspaper,” the puzzle was built so that both electoral outcomes were correct answers, requiring 7 other clues to have dual responses.
