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.
John Went to the nearby store in a Mall to buy something for her home. Below is the conversation between the two:
John: How much for the one?
Shopkeeper: It is $2
John: How much for the Eleven?
Shopkeeper: It is $4
John: How much for the Hundred?
Shopkeeper: It is $6.
I am eight letters long - "12345678"
My 1234 is an atmospheric condition.
My 34567 supports a plant.
My 4567 is too appropriate.
My 45 is a friendly thank-you.
My 678 is a man's name.