Puzzle - 4

Last time I had asked the following two puzzles

An apple vendor has 1000 apples and 10 empty boxes. He asks his son to place all the 1000 apples in the 10 boxes in such a manner that if he asks for any number of apples from 1 to 1000, his son should be able to pick them in terms of boxes - i.e. by picking up one or more full boxes. How did the son place all the apples among the 10 boxes, given that any number of apples can be put in one box?
 There are ‘n’ numbers of boxes. They contain as many balls as one can think of.  Out of these in ‘m’ boxes each balls weighs 1.1 kg. In rest of the boxes that is in ‘n minus m’ boxes balls weigh 1 kg. There is a spring balance. In how many weighing one can find the ‘m’ boxes that have balls weighing 1.1 kg.

One could do it ‘n’ weighing. But that would be waste of time. What is the least number of weighing in which it can be found out?

 The answers to these puzzles are one and the same. For the first puzzle we should keep apples in powers of 2 namely 1 apple in the first box, 2 in the second, 4 in the third, 8 in the fourth and so on. 
As for the second puzzle one can find out the boxes having balls of 1.1 kg in one weighing only.  We should take out the balls from different boxes in powers of two namely, 1, 2, 4, 8, 16, 32 (as we had kept the apples) and weigh them together.  If all boxes had balls weighing 1 Kg then total weight will be 2ⁿ-1 Kg.  For example if there are 10 boxes to begin with and all balls weighed 1 kg. each then total weight  would be 2¹⁰-1 i.e. 1023 Kgs.  In case there are balls weighing 1.1 Kg then there would be excess weight. This excess weight gives a unique combination to find out the boxes that contain balls with 1.1 kg.  For example if the weight comes out to be 1023.9 kg then the first and fourth box contain balls weighting 1.1 kg.   

Continued >----