1. 0x4D4F4444206973204F4B071A

2. 2²⁰ or approximately 10⁶ (about a million)

3. Throughput is about 0.5 MB/sec after deducting for error-correcting overhead. Divide 5000 MB (or, technically, 5120 MB) by 0.5 MB/sec to get 10,000 (or 10,240) seconds, which is about 2.8 hours.

4. which means 160 of the old computers.

     It is also acceptable is to estimate this way: computers. If you do that,

     though, you must round up to get 157 old computers, since a “quarter computer” makes no sense.


5. 13 GHz = 13 billion cycles per second = 13,000,000,000 cycles per second [commas are required]

6.(a) Lead digit is 8 or greater.
   (b) Twos complement is 0x00000147 = 1(16²) + 4(16¹) + 7 = 327, but we must use a negative sign.
       Final answer: –327.

7. –326 = 0xFFFFFEBA

8. add, twos complement, X [note correct spelling of “complement” with no “i”]

9.(a) dword
   (b) 0000 0000 0000 0000 1110 0101 1000 1011
   (c) 14(16³) + 5(16²) + 8(16) + 11 = 58,763

10.(a) First, find twos complement of 0x718AC32D, which is 0x8E753CD3.
       Then, perform a standard addition as follows:

      111
 0x0000E58B
+0x8E753CD3
 0x8E76225E

       However, this answer is negative (lead digit 8 or more), so we must
       compute its complement in order to determine what value is intended.
       The twos complement of 0x8E76225E is 0x7189DDA2, which equals
       7(16⁷) + 1(16⁶) + 8(16⁵) + 9(16⁴) + 13(16³) + 13(16²) + 10(16) + 2 = 1,904,860,578.
       Final answer must be negative: –1,904,860,578.

11. ask-ee
     8, 2
     2¹²⁸ = 2^{10 · 12.8} = (2¹⁰)^12.8 (10³)^12.8 = 10^{3 · 12.8} = 10^38.4 10³⁹
     10th power