JOMO 6, Short 1

We know that every number has a specific representation in Binary form.

For example, $14$ in base 10 is $1110$ in base 2.

Thus we can write $14$ as $2^3+2^2+2^1$ . Like this , because each number has it's $\color{#D61F06}{Specific}$ $\color{#3D99F6}{Representation}$ and thus $\color{#20A900}{\text{each number}}$ can be written as sum of powers of two.

Thus if you want all the numbers till $143$ to be written as sum of boxes' numbers, then you must get the boxes as $2^0 , 2^1,2^3,2^4,...$ until the sum doesn't exceed 143, i.e. you will have to take the powers of 2 till $2^6 =64$ .

Here we use the result $\displaystyle \sum _{k=0} ^n = 2^{n+1}-1$

Thus the sum of diamonds in the boxes till 64 is $127$ .

Hence there are $143-127=16$

Thus the boxes will have $1,2,4,8,16,32,64 \text{ and } 16$ diamonds hence there will be $\boxed{8}$ boxes needed if you are to minimize the number of boxes.

By simple logic we can say that the first box must have 1 diamond and the second should have 2. now in such a way Aditya can give 1 or 2 or 3 diamonds to Sam.therefore the third box should have 4 diamonds.the fourth box should have 8 diamonds and so on.. The total number of boxes required is 8. So the number of diamonds in the boxes will be $1,2,4,8,16,32,64,16$ ... note that 128 won't be there because then the sum of numbers exceeds 143.

Thus , the answer is number of boxes and it is $\boxed{8}$ .

The general solution for N number of diamonds is

x such that $2^{x}$ -1 >=N

use binary