Problem 1: Because data can only be stored in 2 states (on/off, open/closed, etc.).
Problem 2: Main memory is very expensive compared to secondary storage.
Problem 3: Secondary memory is very slow compared to main memory.
Problem 4: (a) Secondary memory (the hard drive)
(b) Main memory (i.e., Random Access Memory)
(c) Secondary memory (the hard drive)
Problem 5: False. Running programs are stored in main memory (i.e., Random Access Memory).
Problem 6: Secondary memory (the hard drive).
Problem 7: The operating system.
Problem 8: (a) 9216 bits / 8 bits per byte = 1152 bytes
(b) 1152 bytes / 4 bytes per word = 288 words
(c) 1152 bytes / 1024 bytes per KB = 1.125 KB
Problem 9. Writing the value of the position below each bit we have
1 0 0 0 0 0
32 16 8 4 2 1
Thus 100000
2

32
10
Problem 10. 10101
2

21
10
.
Problem 11: 0 1 1 1 0 1 0 1
2
7
2
6
2
5
2
4
2
3
2
2
2
1
2
0
0(128) + 1(64) + 1(32) + 1(16) + 0(8) + 1(4) + 0(2) + 1(1) = 64 + 32 + 16 + 4 + 1 = 117
10
Problem 12: 00101110 in base 2 is equivalent to 46 in base 10.
Problem 13. 78
10

1001110
2
Problem 14: 43
10

101011
2
5

Problem 15: 2
8
= 256 and 2
7
= 128 , so there is a 1 in the 2
7
position.
223 – 128 = 95. Since 2
6
= 64, there is a 1 in the 2
6
position.
95 – 64 = 31. Since 2 5 = 32 and 2 4 = 16, there is a 1 in the 2 4 position.
31 – 16 = 15. Since 2
3
= 8, there is a 1 in the 2
3
position.
15 – 8 = 7. Since 2
2
= 4, there is a 1 in the 2
2
position.
7 – 4 = 3. Since 2 1 = 2, there is a 1 in the 2 1 position.
3 – 2 = 1. Since 2
0
= 1, there is a 1 in the 2
0
position.
1 1 0 1 1 1 1 1
2
7
2
6
2
5
2
4
2
3
2
2
2
1
2
0
Problem 16. 0x27 is equivalent to 39 in base-10.
= 11011111
2
Problem 17: 0xBEE is equivalent to 3054 in base-10.
Problem 18. 0x100 is 100000000 in binary.
Problem 19: 0xF1E is equivalent to 111100011110 in binary.
Problem 20: 0xB9B
Problem 21. Since the first hexadecimal digit is C, the first four bits are 1100. Thus, the fourth bit is a zero.
Note how more difficult this question would have been if the address was provided in base-10 instead of in base-16.
Problem 22: The answer is 0x0000006B . The picture is:
Address
06B
06C
C
What is stored
is stored in these two bytes
06D
06E
06F
070
B is stored in these four bytes
071
072
073
A is stored in these four bytes
074
Problem 23: The letter k is stored as ASCII 0x6B which in binary is 01101011 .
Problem 24: The letter g is stored as ASCII 0x67 .
Problem 25: 0x138 + 0x6 = 0x13E .
Problem 26. 0x8D4 .
Problem 27: 0x113D
6

Problem 28:
(a)
0804839d
0804839e
0804839f
080483a0
080483a1
080483a2
080483a3
080483a4
(b) The four preceding addresses are (from lowest to highest):
08048398
08048399
0804839a
0804839b
(c) In ASCII, 0x55 = U, 0x53 =S, 0x4E = N, 0x41 = A, 0x21 = ! Thus: USNA!
Problem 29. 16 into 730 gives a quotient of 45 with a remainder of A
16 into 45 gives a quotient of 2 with a remainder of D
16 into 2 gives a quotient of 0 with a remainder of 2
Thus, 730
10

0x2DA . Since we are asked to use the number of hex digits appropriate for the x86 architecture, we must use 8 hex digits. Thus, the final answer is 0x000002DA
7