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.
You have four chains. Each chain has three links in it. Although it is difficult to cut the links, you wish to make a single loop with all 12 links. What is the fewest number of cuts you must make to accomplish this task?
Consider this about a word: The first two letters signify a male, the first three letters signify a female, the first four letters signify a great, while the entire world signifies a great woman. What is the word?
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.
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