End-of-Unit Problems: Number Systems

Work through these problems to reinforce your understanding of number systems and conversions.

Section A: Base Conversions (5 problems)

Problem 1

Convert the following decimal numbers to binary, octal, and hexadecimal:

a) 156 b) 243 c) 512 d) 1000

Show Solution

a) 156₁₀ = 10011100₂ = 234₈ = 9C₁₆

b) 243₁₀ = 11110011₂ = 363₈ = F3₁₆

c) 512₁₀ = 1000000000₂ = 1000₈ = 200₁₆

d) 1000₁₀ = 1111101000₂ = 1750₈ = 3E8₁₆

Problem 2

Convert the following binary numbers to decimal:

a) 11010110₂ b) 10101010₂ c) 11111111₂ d) 10000001₂

Show Solution

a) 11010110₂ = 128 + 64 + 16 + 4 + 2 = 214₁₀

b) 10101010₂ = 128 + 32 + 8 + 2 = 170₁₀

c) 11111111₂ = 255₁₀ (all bits set)

d) 10000001₂ = 128 + 1 = 129₁₀

Problem 3

Convert between hexadecimal and binary:

a) 3F7₁₆ to binary b) 1010111001₂ to hexadecimal c) BEEF₁₆ to binary d) 110011110000₂ to hexadecimal

Show Solution

a) 3F7₁₆ = 0011 1111 0111₂ = 001111110111₂

b) 1010111001₂ = 0010 1011 1001₂ = 2B9₁₆

c) BEEF₁₆ = 1011 1110 1110 1111₂ = 1011111011101111₂

d) 110011110000₂ = 1100 1111 0000₂ = CF0₁₆

Problem 4

Convert the following octal numbers to decimal and binary:

a) 752₈ b) 377₈ c) 1234₈ d) 4000₈

Show Solution

a) 752₈ = 7×64 + 5×8 + 2×1 = 448 + 40 + 2 = 490₁₀ = 111101010₂

b) 377₈ = 3×64 + 7×8 + 7×1 = 192 + 56 + 7 = 255₁₀ = 11111111₂

c) 1234₈ = 1×512 + 2×64 + 3×8 + 4×1 = 512 + 128 + 24 + 4 = 668₁₀

d) 4000₈ = 4×512 = 2048₁₀ = 100000000000₂

Problem 5

A computer memory address is given as 7FFF₁₆.

a) Convert this to decimal b) Convert this to binary c) How many bytes can be addressed from 0000₁₆ to 7FFF₁₆?

Show Solution

a) 7FFF₁₆ = 7×4096 + 15×256 + 15×16 + 15 = 28672 + 3840 + 240 + 15 = 32767₁₀

b) 7FFF₁₆ = 0111 1111 1111 1111₂ = 111111111111111₂ (15 ones)

c) From 0000 to 7FFF = 7FFF + 1 = 8000₁₆ = 32768 bytes = 32 KB

Section B: Signed Number Representations (5 problems)

Problem 6

Represent -45 in the following 8-bit formats:

a) Sign-magnitude b) One's complement c) Two's complement

Show Solution

First, 45₁₀ = 00101101₂

a) Sign-magnitude: 10101101 (MSB = 1 for negative)

b) One's complement: Invert all bits of 00101101 → 11010010

c) Two's complement: One's complement + 1 = 11010010 + 1 = 11010011

Problem 7

What decimal values do the following 8-bit two's complement numbers represent?

a) 10000000 b) 11111111 c) 10000001 d) 01111111

Show Solution

a) 10000000₂ = -128 (most negative 8-bit value)

b) 11111111₂ = -1 (invert → 00000000, add 1 → 00000001, negate → -1)

c) 10000001₂ = -127 (invert → 01111110 = 126, add 1 → 127, negate → -127)

d) 01111111₂ = +127 (MSB = 0, so positive, value = 127)

Problem 8

What is the range of values that can be represented in:

a) 8-bit unsigned b) 8-bit two's complement c) 16-bit unsigned d) 16-bit two's complement

Show Solution

a) 8-bit unsigned: 0 to 255 (0 to \(2^8 - 1\))

b) 8-bit two's complement: -128 to +127 (\(-2^7\) to \(2^7 - 1\))

c) 16-bit unsigned: 0 to 65,535 (0 to \(2^{16} - 1\))

d) 16-bit two's complement: -32,768 to +32,767 (\(-2^{15}\) to \(2^{15} - 1\))

Problem 9

Perform the following two's complement arithmetic (8-bit):

a) 35 + 47 b) 35 + (-47) c) (-35) + (-47) d) 100 + 50 (check for overflow)

Show Solution

a) 35 = 00100011, 47 = 00101111
00100011 + 00101111 = 01010010 = 82 ✓

b) 35 = 00100011, -47 = 11010001
00100011 + 11010001 = 11110100 = -12 ✓

c) -35 = 11011101, -47 = 11010001
11011101 + 11010001 = 110101110 (carry out)
Result = 10101110 = -82 ✓

d) 100 = 01100100, 50 = 00110010
01100100 + 00110010 = 10010110 = -106
Overflow! (positive + positive = negative)
Actual answer should be 150, but it exceeds +127

Problem 10

Explain why overflow occurred in Problem 9d and how you can detect it.

Show Solution

Why overflow occurred:

● Both operands (100 and 50) are positive
● The true sum (150) exceeds the maximum positive value (+127) for 8-bit two's complement
● The result wrapped around to a negative number (-106)

Overflow detection rules:

● Overflow occurs when adding two positive numbers yields a negative result
● Overflow occurs when adding two negative numbers yields a positive result
● Overflow CANNOT occur when adding numbers of opposite signs

Hardware detection: Overflow = Carry into MSB XOR Carry out of MSB.
In this case: Carry in = 1, Carry out = 0, so 1 XOR 0 = 1 (overflow)

Section C: Binary Arithmetic (3 problems)

Problem 11

Perform binary addition:

a) 10110 + 11011 b) 111111 + 000001 c) 10101010 + 01010101

Show Solution

a)

    10110
  + 11011
  -------
   110001

22 + 27 = 49 ✓

b)

    111111
  + 000001
  --------
   1000000

63 + 1 = 64 ✓

c)

    10101010
  + 01010101
  ----------
    11111111

170 + 85 = 255 ✓

Problem 12

Perform binary subtraction using two's complement:

a) 11010 - 10011 b) 10000 - 00001 c) 01100100 - 00110010

Show Solution

a) 11010 - 10011 = 11010 + (two's complement of 10011)
Two's complement of 10011 = 01101
11010 + 01101 = 100111 → discard carry → 00111 = 7
Check: 26 - 19 = 7 ✓

b) 10000 - 00001 = 10000 + 11111 = 101111 → 01111 = 15
Check: 16 - 1 = 15 ✓

c) 01100100 - 00110010 = 01100100 + 11001110 = 100110010
Discard carry → 00110010 = 50
Check: 100 - 50 = 50 ✓

Problem 13

Multiply the following binary numbers:

a) 1101 × 101 b) 1011 × 110

Show Solution

a) 1101 × 101:

        1101
      ×  101
      ------
        1101  (1101 × 1)
       0000   (1101 × 0, shifted)
      1101    (1101 × 1, shifted)
      ------
     1000001

13 × 5 = 65 = 1000001₂ ✓

b) 1011 × 110:

        1011
      ×  110
      ------
       0000   (1011 × 0)
      1011    (1011 × 1, shifted)
     1011     (1011 × 1, shifted)
     -------
     1000010

11 × 6 = 66 = 1000010₂ ✓

Section D: Fractional Numbers (2 problems)

Problem 14

Convert the following decimal fractions to binary (4 fractional bits):

a) 0.625 b) 0.3125 c) 6.75

Show Solution

a) 0.625 × 2 = 1.25 → 1
0.25 × 2 = 0.5 → 0
0.5 × 2 = 1.0 → 1
0.625₁₀ = 0.101₂

b) 0.3125 × 2 = 0.625 → 0
0.625 × 2 = 1.25 → 1
0.25 × 2 = 0.5 → 0
0.5 × 2 = 1.0 → 1
0.3125₁₀ = 0.0101₂

c) Integer part: 6 = 110₂
Fractional: 0.75 × 2 = 1.5 → 1, 0.5 × 2 = 1.0 → 1
6.75₁₀ = 110.11₂

Problem 15

Convert the following binary fractions to decimal:

a) 101.011₂ b) 11.1101₂ c) 0.0001₂

Show Solution

a) 101.011₂ = 4 + 0 + 1 + 0.25 + 0.125 = 5.375₁₀

b) 11.1101₂ = 2 + 1 + 0.5 + 0.25 + 0 + 0.0625 = 3.8125₁₀

c) 0.0001₂ = 0.0625₁₀ = 1/16

Section E: Application Problems (5 problems)

Problem 16

An 8-bit grayscale image uses values 0-255 to represent pixel brightness.

a) What binary value represents 50% gray? b) What is the hex value for white? c) How many distinct shades can be represented?

Show Solution

a) 50% gray = 255 × 0.5 ≈ 127 or 128
127 = 01111111₂ or 128 = 10000000₂

b) White = maximum brightness = 255 = FF₁₆

c) 8 bits = 2⁸ = 256 distinct shades (0 to 255)

Problem 17

A 24-bit RGB color uses 8 bits each for Red, Green, and Blue.

a) What is the hex representation of pure red? b) What color is #00FF00? c) What color is #808080? d) How many colors can be represented?

Show Solution

a) Pure red: R=255, G=0, B=0 → #FF0000

b) R=0, G=255, B=0 → Pure green

c) R=128, G=128, B=128 → Medium gray (50% gray)

d) 24 bits = 2²⁴ = 16,777,216 colors

Problem 18

A 4-bit binary counter counts from 0000 to 1111 repeatedly.

a) How many states does it have? b) If it increments every 1 ms, how long until it returns to 0000? c) What is the state after 0111? d) What is the state after 1111?

Show Solution

a) 4 bits = 2⁴ = 16 states (0 to 15)

b) 16 states × 1 ms = 16 ms for one complete cycle

c) 0111 + 1 = 1000 (7 → 8)

d) 1111 + 1 = 10000, but only 4 bits, so 0000 (wrap around)

Problem 19

A computer uses 32-bit addresses.

a) How many bytes of memory can be addressed? b) Express this in KB, MB, and GB c) If upgrading to 64-bit addressing, how much more memory can be addressed?

Show Solution

a) \(2^{32}\) = 4,294,967,296 bytes

b) 4,294,967,296 bytes = 4,194,304 KB = 4,096 MB = 4 GB

c) 64-bit: \(2^{64}\) = 18,446,744,073,709,551,616 bytes = 16 exabytes
Increase factor: \(2^{64} / 2^{32} = 2^{32}\) = 4 billion times more

Problem 20

Design a number format for a simple embedded system that needs to represent:

Temperatures from -40°C to +85°C
Resolution of 0.5°C

a) How many bits are needed? b) What encoding scheme would you use? c) Show the encoding for -10°C, 0°C, and 25.5°C

Show Solution

a) Range: -40 to +85 = 126 degrees
With 0.5°C resolution: 126 / 0.5 = 252 values needed
\(2^8 = 256\), so 8 bits are sufficient

b) Offset binary encoding:

● Offset = 40 (to make -40°C map to 0)
● Value = (Temperature + 40) / 0.5
● Or use two's complement with implicit scaling

c) Using offset binary (offset = 40, scale = 0.5):

● -10°C: (-10 + 40) / 0.5 = 60 = 00111100₂
● 0°C: (0 + 40) / 0.5 = 80 = 01010000₂
● 25.5°C: (25.5 + 40) / 0.5 = 131 = 10000011₂

Summary

Section	Topics Covered	Problem Count
A	Base Conversions	5
B	Signed Numbers	5
C	Binary Arithmetic	3
D	Fractional Numbers	2
E	Applications	5
Total		20