# Number System - An Introduction

## Tick (✓) the correct answer

#### Question 1

In a decimal number system, the base of a number is represented by

1. 2
2. 10 ✓
3. 16
4. All of the above

#### Question 2

The base of an octal number is represented by:

1. 2
2. 8 ✓
3. 7
4. None

#### Question 3

To convert an octal number to its binary equivalent, each octal digit is expressed as

1. 3 bits form ✓
2. 4 bits form
3. 8 bits form
4. All of the above

#### Question 4

Sixteen raised to the power zero (16⁰) is equivalent to

1. 0
2. 1 ✓
3. 0 and 1
4. None

#### Question 5

An octal number system uses the digits from

1. 0 to 8
2. 1 to 8
3. 0 to 7 ✓
4. All of the above

#### Question 6

The base of a hexa-decimal number is represented by

1. H16
2. 16 ✓
3. 15
4. None

#### Question 7

In a hexa-decimal number system, 'B' represents the digit

1. 11 ✓
2. 12
3. 14
4. 13

#### Question 8

To express a hexa-decimal number to its binary equivalent, each hexa-decimal digit is expressed as

1. 2 bits form
2. 3 bits form
3. 4 bits form ✓
4. None

#### Question 9

The binary equivalent of a hexa-decimal digit 12(C) is represented by

1. 1010
2. 1011
3. 1101
4. 1100 ✓

#### Question 10

The hexa-decimal equivalent digit of 1011 (4 bits form) is

1. 14
2. 15
3. 11 ✓
4. 12

## Fill in the blanks

#### Question 1

The binary system consists of two digits 0 and 1.

#### Question 2

A decimal number system uses the digits from 0 to 9.

#### Question 3

The base in a decimal number system is written as 10.

#### Question 4

A binary number system is written with 2 as the base.

#### Question 5

In a decimal to binary conversion, the first remainder is known as Least Significant Bit (LSB) and the last remainder is Most Significant Bit (MSB).

20 = 1

Octal
Digit
Binary
Equivalent
5
7
1
6
3
Digit
Binary
Equivalent
8
11
4
15
9

Octal
Digit
Binary
Equivalent
5101
7111
1001
6110
3011
Digit
Binary
Equivalent
81000
111011
40100
151111
91001

## Convert the following to their binary equivalents

#### Question 1

(78)10

2QuotientRemainder
2780 (LSB)
2391
2191
291
240
220
211 (MSB)
0

Therefore, (78)10 = (1001110)2

#### Question 2

(99)10

2QuotientRemainder
2991 (LSB)
2491
2240
2120
260
231
211 (MSB)
0

Therefore, (99)10 = (1100011)2

#### Question 3

(141)10

2QuotientRemainder
21411 (LSB)
2700
2351
2171
280
240
220
211 (MSB)
0

Therefore, (141)10 = (10001101)2

#### Question 4

(123)10

2QuotientRemainder
21231 (LSB)
2611
2300
2151
271
231
211 (MSB)
0

Therefore, (123)10 = (1111011)2

## Convert the following to their decimal equivalents

#### Question 1

(10101)2 to ( )10

Binary
No
PowerValueResult
1 (LSB)2011x1=1
02120x2=0
12241x4=4
02380x8=0
1 (MSB)24161x16=16

Equivalent decimal number = 1 + 4 + 16 = 21

Therefore, (10101)2 = (21)10

#### Question 2

(10000)2 to ( )10

Binary
No
PowerValueResult
0 (LSB)2010x1=0
02120x2=0
02241x4=4
02380x8=0
1 (MSB)24161x16=16

Equivalent decimal number = 16

Therefore, (10000)2 = (16)10

#### Question 3

(11001)2 to ( )10

Binary
No
PowerValueResult
1 (LSB)2011x1=1
02120x2=0
02240x4=0
12381x8=8
1 (MSB)24161x16=16

Equivalent decimal number = 1 + 8 + 16 = 25

Therefore, (11001)2 = (25)10

#### Question 4

(101010)2 to ( )10

Binary
No
PowerValueResult
0 (LSB)2010x1=0
12121x2=2
02240x4=0
12381x8=8
024160x16=0
1 (MSB)25321x32=32

Equivalent decimal number = 2 + 8 + 32 = 42

Therefore, (101010)2 = (42)10

## Convert the following to Decimal numbers

#### Question 1

(510)8

Octal
No
PowerValueResult
0 (LSB)8010x1=0
18181x8=8
5 (MSB)82645x64=320

Equivalent decimal number = 8 + 320 = 328

Therefore, (510)8 = (328)10

#### Question 2

(ABC)16

Number
PowerValueResult
C (12)160112x1=12
B (11)1611611x16=176
A (10)16225610x256=2560

Equivalent decimal number = 12 + 176 + 2560 = 2748

Therefore, (ABC)16 = (2748)10

#### Question 3

(1001011)2

Binary
No
PowerValueResult
1 (LSB)2011x1=1
12121x2=2
02240x4=0
12381x8=8
024160x16=0
025320x32=0
1 (MSB)26641x64=64

Equivalent decimal number = 1 + 2 + 8 + 64 = 75

Therefore, (1001011)2 = (75)10

#### Question 4

(CD7)16

Number
PowerValueResult
716017x1=7
D (13)1611613x16=208
C (12)16225612x256=3072

Equivalent decimal number = 7 + 208 + 3072 = 3287

Therefore, (CD7)16 = (3287)10

#### Question 5

(101001)2 to ( )10

Binary
No
PowerValueResult
1 (LSB)2011x1=1
02120x2=0
02240x4=0
12381x8=8
024160x16=0
1 (MSB)25321x32=32

Equivalent decimal number = 1 + 8 + 32 = 41

Therefore, (101001)2 = (41)10

#### Question 6

(1100111)2 to ( )10

Binary
No
PowerValueResult
1 (LSB)2011x1=1
12121x2=2
12241x4=4
02380x8=0
024160x16=0
125321x32=32
1 (MSB)26641x64=64

Equivalent decimal number = 1 + 2 + 4 + 32 + 64 = 103

Therefore, (1100111)2 = (103)10

## Perform the following

#### Question 1

(342)8 to ( )2

Octal
Number
Binary
Equivalent
2010
4100
3011

Therefore, (342)8 = ($\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{100}}\medspace\bold{\underlinesegment{010}}$)2

#### Question 2

(203)8 to ( )2

Octal
Number
Binary
Equivalent
3011
0000
2010

Therefore, (203)8 = ($\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{011}}$)2

#### Question 3

Number
Binary
Equivalent
D (13)1101
A (10)1010
91001

Therefore, (9AD)16 = ($\bold{\underlinesegment{1001}}\medspace\bold{\underlinesegment{1010}}\medspace\bold{\underlinesegment{1101}}$)2

#### Question 4

(157)8 to ( )2

Octal
Number
Binary
Equivalent
7111
5101
1001

Therefore, (157)8 = ($\bold{\underlinesegment{001}}\medspace\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{111}}$)2

#### Question 5

(ABC)16 to ( )2

Number
Binary
Equivalent
C (12)1100
B (11)1011
A (10)1010

Therefore, (ABC)16 = ($\bold{\underlinesegment{1010}}\medspace\bold{\underlinesegment{1011}}\medspace\bold{\underlinesegment{1100}}$)2

#### Question 6

(DE)16 to ( )2

Number
Binary
Equivalent
E (14)1110
D (13)1101

Therefore, (DE)16 = ($\bold{\underlinesegment{1101}}\medspace\bold{\underlinesegment{1110}}$)2

## Convert the following to their hexa-decimal equivalent

#### Question 1

(110011101111)2

Grouping in bits of 4:

$\underlinesegment{1100} \quad \underlinesegment{1110} \quad \underlinesegment{1111}$

Binary
Number
Equivalent
1111F (15)
1110E (14)
1100C (12)

Therefore, (110011101111)2 = (CEF)16

#### Question 2

(11010111100)2

Grouping in bits of 4:

$\underlinesegment{0110} \quad \underlinesegment{1011} \quad \underlinesegment{1100}$

Binary
Number
Equivalent
1100C (12)
1011B (11)
01106

Therefore, (11010111100)2 = (6BC)16

#### Question 3

(89392)10

16QuotientRemainder
16893920
1655873
16349D (13)
16215
1611
0

Therefore, (89392)10 = (15D30)16

#### Question 4

(100101101110)2

Grouping in bits of 4:

$\underlinesegment{1001} \quad \underlinesegment{0110} \quad \underlinesegment{1110}$

Binary
Number
Equivalent
1110E (14)
01106
10019

Therefore, (100101101110)2 = (96E)16

#### Question 5

(9894)10

16QuotientRemainder
1698946
16618A (10)
16386
1622
0

Therefore, (9894)10 = (26A6)16

#### Question 6

(4966)10

16QuotientRemainder
1649666
163106
16193
1611
0

Therefore, (4966)10 = (1366)16

#### Question 1

What are the different types of number systems that a computer deals with?

The different types of number systems are:

1. Binary Number System
2. Octal Number System
3. Decimal Number System

#### Question 2

What is meant by the following terms? Give an example of each.

(a) An octal number
(b) A hexa-decimal number

(a) An Octal number — An octal number uses 8 types of digits — 0, 1, 2, 3, 4, 5, 6, 7. It is represented with base 8.

(b) A hexa-decimal number — A Hexa-decimal number uses 16 types of digits (0 to 15). To represent digits from 10 to 15 it uses letters from A to F respectively. It is represented with base 16.

#### Question 3a

Give two differences between Binary number and Decimal number

Binary numberDecimal number
It uses 2 digits — 0 and 1.It uses 10 digits — 0 to 9.
It uses base 2.It uses base 10.

#### Question 3b

Give two differences between Octal number and Binary number