* Exercise #1:

  - Question #1:
        If a block pointer uses 2^3 bytes, and a block is 2^10 bytes, then each block can hold 2^7 pointers
        Therefore, an i-node with oly the single-indirect can describe a file with 2^3 + 2^7 blocks

  - Question #2:
        If a block pointer uses 2^3 bytes, and a block is 2^10 bytes, then each block can hold 2^7 pointers
        Therefore, an i-node that uses all its pointers fully can describe a file with 
            2^3 + 2^7 + 2^14 + 2^21 blocks

  - Question #3: 
        The file itself will use 33 blocks. The i-node is on one block. There needs to be an extra block for a single-indirect
        that will point to 33-8 = 25 blocks  (no need for double indirect since with the single indirect we can address 2^7 blocks,
        way more than 25).  Answer: 1 + 1 + 33 = 35

  - Question #4:
        A 1GiB file needs 2^20 blocks
        Using the direct, single-, and double- indirect to their fullest,
        and assuming X is the block pointer size in byte, we have:

        2^20 = 8 + (2^10 / X)  + (2^10 / X) * (2^10 / X)
        
        where the first term is the number of blocks accessible via direct pointers
              the second term is the number of blocks accessible via single-indirect pointers
              the third term is the number of blocks accessible via double-indirect pointers

        The above equation can be rewritten as:

        2^10 X^2 =  8 X^2 + X + 2^10
        
        of   1016 X^2 - X - 1024 = 0

        This is a second-degree equation with solution:  X = 1.004..  which is just above 1 byte

        Therefore, pointers are 9 bits