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.
By using all numbers, i.e. 123456789 and subtraction/addition, operators number 100 can be formed in many ways.
Example: 98 + 7 + 6 - 5 - 4 - 3 + 2 - 1 = 100
But if we add a condition use of the number 32 is a must. Then there are limited solutions.
One of such solution is: 9 - 8 + 76 + 54 - 32 + 1 = 100
A rubber ball keeps on bouncing back to 2/3 of the height from which it is dropped. Can you calculate the fraction of its original height that the ball will bounce after it is dropped and it has bounced four times without any hindrance ?