SOLUTIONS TO HOMEWORK PROBLEMS

Chapter 1

1.3    a.    21 = 2
         b.    23 = 8
         c.    26 = 64
         d.    28 = 256
         e.    210 = 1024
         f.    216 = 65536

1.5    8 because 28 = 256 the smallest power of 2 greater than 240

1.6    1 kilobyte = 1024 bytes = 1024 * 8 = 8192 bits.  So
         a.    12 KB = 12 * 8192 = 98304 bits
         1 megabyte = 1024*1024 = 1048576 bytes = 8388608 bits
         b.    5 MG = 5*8388608 = 41943040 bits
        1 gigabyte = 1073741824 bytes = 8589934592 bits
         c.    3 GB = 25769803776 bits
        1 terabyte = 109951162776 bytes = 8796093022208 bits
        d.    2 TB = 17592186044416 bits

1.16.    The invalid identifiers are 2ndLevel (starts with a number) and hook&ladder (& is not a Java letter or digit)
            The rest are valid although anExtremelyLongIdentifierIfYouAskMe is unnecessarily long

1.17    q does not give sufficient information
           totVal can be confusing
           theNextValueInTheList is long and unnecessary.  nextValue gives the same information
 
1.22    The total number of pixels is 400*250 = 100000 pixels
           Each pixel requires 1 Byte or 8 bits.  So the total number of bits is 800000.