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.
In the picture, you can see a chess board. On the top left position, the K marks a knight. Now, can you move the knight in a manner that after 63 moves, the knight has been placed at all the squares exactly once excluding the starting square?
There is a shop where written:
Buy 1 for $1
10 for $2
100 for $3
I needed 999 and still only paid $3. How could this be financially viable for the shop-keeper?
In 2011, people playing Foldit, an online puzzle game about protein folding, resolved the structure of an enzyme that causes an Aids-like disease in monkeys. Researchers had been working on the problem for 13 years. The gamers solved it in three weeks.
