I am thinking of a five-digit number such that:
The first and last digits are the same, their submission is an even number and multiplication is an odd number and is equal to the fourth number. Subtract five from it and we obtain the second number. Then divide into exact halves and we get the 3rd number.
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.