Chapter 2 WHOLE NUMBERS
BINARY NUMBER SYSTEM
Quinary NUMBER SYSTEM
1. Define the following:
(a) Decimal Number System (Base-10).
The decimal number system is a base-10 numbering system that uses ten distinct digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Each position in a decimal number represents a power of 10.
(b) Binary Number System (Base-2).
The binary number system is a base-2 numbering system that uses only two digits: 0 and 1. Each position in a binary number represents a power of 2.
2. What are the digits used in binary number?
The digits used in the binary number system are 0 and 1.
3. How many digits are used in decimal system?
The decimal system uses 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
4. Write the binary number equal to ( 1 x 2² + 1 x 2¹ + 1 x 2⁰ ).
Solution:
1 x 2² + 1 x 2¹ + 1 x 2⁰
= 111₂
5. What is the binary representation of the decimal number 3?
3 = 2 + 1 = 1 x 2¹ + 1 x 2⁰ = 11₂
6. Define the Quinary Number System (Base-5).
The quinary number system is a base-5 numbering system that uses five distinct digits: 0, 1, 2, 3, and 4. Each position in a quinary number represents a power of 5.
7. How many digits are used in quinary system?
The quinary system uses 5 digits: 0, 1, 2, 3, and 4.
8. Write the quinary number equal to ( 4 x 5³ + 3 x 5² + 0 x 5¹ + 2 x 5⁰ ).
Solution:
4 x 5³ + 3 x 5² + 0 x 5¹ + 2 x 5⁰
= 4302₅
9. What is the quinary representation of the decimal number 6?
= 5 + 1
= 1 x 5¹ + 1 x 5⁰
= 11₅
The quinary representation is 11.
SOLVE
1. Write the following decimal numbers in expanded form:
(a) 242
Solution:
= 200 + 40 + 2
= 2×10² + 4×10¹ + 2×10⁰
(b) 3941
Solution:
= 3000 + 900 + 40 + 1
= 3×10³ + 9×10² + 4×10¹ + 1×10⁰
(c) 4346
Solution:
= 4000 + 300 + 40 + 6
= 4×10³ + 3×10² + 4×10¹ + 6×10⁰
(d) 50391
Solution:
= 50000 + 0 + 300 + 90 + 1
= 5×10⁴ + 0×10³ + 3×10² + 9×10¹ + 1×10⁰
(e) 46921
Solution:
= 40000 + 6000 + 900 + 20 + 1
= 4×10⁴ + 6×10³ + 9×10² + 2×10¹ + 1×10⁰
(f) 792311
Solution:
= 700000 + 90000 + 2000 + 300 + 10 + 1
= 7×10⁵ + 9×10⁴ + 2×10³ + 3×10² + 1×10¹ + 1×10⁰
(g) 246321
Solution:
= 200000 + 40000 + 6000 + 300 + 20 + 1
= 2×10⁵ + 4×10⁴ + 6×10³ + 3×10² + 2×10¹ + 1×10⁰
(h) 5913217
Solution:
= 5000000 + 900000 + 10000 + 3000 + 200 + 10 + 7
= 5×10⁶ + 9×10⁵ + 1×10⁴ + 3×10³ + 2×10² + 1×10¹ + 7×10⁰
2. Convert the following in index form of 2:
(a) 64
Solution:
64 = 1 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 0 × 20
(b) 90
Solution:
90 = 1 × 26 + 0 × 25 + 1 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20
(c) 104
Solution:
104 = 1 × 26 + 1 × 25 + 0 × 24 + 1 × 23 + 0 × 22 + 0 × 21 + 0 × 20
(d) 120
Solution:
120 = 1 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 0 × 22 + 0 × 21 + 0 × 20
(e) 240
Solution:
240 = 1 × 27 + 1 × 26 + 1 × 25 + 1 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 0 × 20
(f) 300
Solution:
300 = 1 × 28 + 0 × 27 + 0 × 26 + 1 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20
(g) 360
Solution:
360 = 1 × 28 + 0 × 27 + 1 × 26 + 1 × 25 + 0 × 24 + 1 × 23 + 0 × 22 + 0 × 21 + 0 × 20
(h) 516
Solution:
516 = 1 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20
3. Convert the following in index form of 5:
(a) 50
Solution:
50 = 2 × 52 + 0 × 51 + 0 × 50
(b) 65
Solution:
65 = 2 × 52 + 3 × 51 + 0 × 50
(c) 95
Solution:
95 = 3 × 52 + 4 × 51 + 0 × 50
(d) 120
Solution:
120 = 4 × 52 + 4 × 51 + 0 × 50
(e) 165
Solution:
165 = 1 × 53 + 0 × 52 + 3 × 51 + 0 × 50
(f) 204
Solution:
204 = 1 × 53 + 3 × 52 + 0 × 51 + 4 × 50
(g) 640
Solution:
640 = 1 × 54 + 0 × 53 + 0 × 52 + 3 × 51 + 0 × 50
(h) 1250
Solution:
1250 = 2 × 54 + 0 × 53 + 0 × 52 + 0 × 51 + 0 × 50
4. Show the following binary number into a place value chart.
(e) 100011₂
Solution:
1 = 1 × 2⁰ = 1
1 = 1 × 2¹ = 2
0 = 0 × 2² = 0
0 = 0 × 2³ = 0
0 = 0 × 2⁴ = 0
1 = 1 × 2⁵ = 32
Decimal Equivalent = 1 + 2 + 0 + 0 + 0 + 32 = 35
(f) 100110₂
Solution:
0 = 0 × 2⁰ = 0
1 = 1 × 2¹ = 2
1 = 1 × 2² = 4
0 = 0 × 2³ = 0
0 = 0 × 2⁴ = 0
1 = 1 × 2⁵ = 32
Decimal Equivalent = 0 + 2 + 4 + 0 + 0 + 32 = 38
(g) 101000₂
Solution:
0 = 0 × 2⁰ = 0
0 = 0 × 2¹ = 0
0 = 0 × 2² = 0
1 = 1 × 2³ = 8
0 = 0 × 2⁴ = 0
1 = 1 × 2⁵ = 32
Decimal Equivalent = 0 + 0 + 0 + 8 + 0 + 32 = 40
(h) 101010₂
Solution:
0 = 0 × 2⁰ = 0
1 = 1 × 2¹ = 2
0 = 0 × 2² = 0
1 = 1 × 2³ = 8
0 = 0 × 2⁴ = 0
1 = 1 × 2⁵ = 32
Decimal Equivalent = 0 + 2 + 0 + 8 + 0 + 32 = 42
(i) 101100₂
Solution:
0 = 0 × 2⁰ = 0
0 = 0 × 2¹ = 0
1 = 1 × 2² = 4
1 = 1 × 2³ = 8
0 = 0 × 2⁴ = 0
1 = 1 × 2⁵ = 32
Decimal Equivalent = 0 + 0 + 4 + 8 + 0 + 32 = 44
(j) 101101₂
Solution:
1 = 1 × 2⁰ = 1
0 = 0 × 2¹ = 0
1 = 1 × 2² = 4
1 = 1 × 2³ = 8
0 = 0 × 2⁴ = 0
1 = 1 × 2⁵ = 32
Decimal Equivalent = 1 + 0 + 4 + 8 + 0 + 32 = 45
(k) 110000₂
Solution:
0 = 0 × 2⁰ = 0
0 = 0 × 2¹ = 0
0 = 0 × 2² = 0
0 = 0 × 2³ = 0
1 = 1 × 2⁴ = 16
1 = 1 × 2⁵ = 32
Decimal Equivalent = 0 + 0 + 0 + 0 + 16 + 32 = 48
(ζ) 110010₂
Solution:
0 = 0 × 2⁰ = 0
1 = 1 × 2¹ = 2
0 = 0 × 2² = 0
0 = 0 × 2³ = 0
1 = 1 × 2⁴ = 16
1 = 1 × 2⁵ = 32
Decimal Equivalent = 0 + 2 + 0 + 0 + 16 + 32 = 50
(m) 110110₂
Solution:
0 = 0 × 2⁰ = 0
1 = 1 × 2¹ = 2
1 = 1 × 2² = 4
0 = 0 × 2³ = 0
1 = 1 × 2⁴ = 16
1 = 1 × 2⁵ = 32
Decimal Equivalent = 0 + 2 + 4 + 0 + 16 + 32 = 54
(n) 111000₂
Solution:
0 = 0 × 2⁰ = 0
0 = 0 × 2¹ = 0
0 = 0 × 2² = 0
1 = 1 × 2³ = 8
1 = 1 × 2⁴ = 16
1 = 1 × 2⁵ = 32
Decimal Equivalent = 0 + 0 + 0 + 8 + 16 + 32 = 56
(o) 111100₂
Solution:
0 = 0 × 2⁰ = 0
0 = 0 × 2¹ = 0
1 = 1 × 2² = 4
1 = 1 × 2³ = 8
1 = 1 × 2⁴ = 16
1 = 1 × 2⁵ = 32
Decimal Equivalent = 0 + 0 + 4 + 8 + 16 + 32 = 60
(p) 111110₂
Solution:
0 = 0 × 2⁰ = 0
1 = 1 × 2¹ = 2
1 = 1 × 2² = 4
1 = 1 × 2³ = 8
1 = 1 × 2⁴ = 16
1 = 1 × 2⁵ = 32
Decimal Equivalent = 0 + 2 + 4 + 8 + 16 + 32 = 62
5. Convert the following binary numbers into decimal number system:
(a) 11001₂
Solution:
11001₂ = 1×2⁴ + 1×2³ + 0×2² + 0×2¹ + 1×2⁰
= 16 + 8 + 0 + 0 + 1
= 25
∴ 11001₂ = 25
(b) 11100₂
Solution:
11100₂ = 1×2⁴ + 1×2³ + 1×2² + 0×2¹ + 0×2⁰
= 16 + 8 + 4 + 0 + 0
= 28
∴ 11100₂ = 28
(c) 11101₂
Solution:
11101₂
= 1×2⁴ + 1×2³ + 1×2² + 0×2¹ + 1×2⁰
= 16 + 8 + 4 + 0 + 1
= 29
∴ 11101₂ = 29
(d) 100000₂
Solution:
100000₂
= 1×2⁵ + 0×2⁴ + 0×2³ + 0×2² + 0×2¹ + 0×2⁰
= 32 + 0 + 0 + 0 + 0 + 0
= 32
∴ 100000₂ = 32
(e) 100011₂
Solution:
100011₂
= 1×2⁵ + 0×2⁴ + 0×2³ + 0×2² + 1×2¹ + 1×2⁰
= 32 + 0 + 0 + 0 + 2 + 1
= 35
∴ 100011₂ = 35
(f) 100110₂
Solution:
100110₂
= 1×2⁵ + 0×2⁴ + 0×2³ + 1×2² + 1×2¹ + 0×2⁰
= 32 + 0 + 0 + 4 + 2 + 0
= 38
∴ 100110₂ = 38
(g) 101000₂
Solution:
101000₂
= 1×2⁵ + 0×2⁴ + 1×2³ + 0×2² + 0×2¹ + 0×2⁰
= 32 + 0 + 8 + 0 + 0 + 0
= 40
∴ 101000₂ = 40
(h) 101010₂
Solution:
101010₂
= 1×2⁵ + 0×2⁴ + 1×2³ + 0×2² + 1×2¹ + 0×2⁰
= 32 + 0 + 8 + 0 + 2 + 0
= 42
∴ 101010₂ = 42
(i) 101100₂
Solution:
101100₂
= 1×2⁵ + 0×2⁴ + 1×2³ + 1×2² + 0×2¹ + 0×2⁰
= 32 + 0 + 8 + 4 + 0 + 0
= 44
∴ 101100₂ = 44
(j) 101101₂
Solution:
101101₂
= 1×2⁵ + 0×2⁴ + 1×2³ + 1×2² + 0×2¹ + 1×2⁰
= 32 + 0 + 8 + 4 + 0 + 1
= 45
∴ 101101₂ = 45
(k) 110000₂
Solution:
110000₂
= 1×2⁵ + 1×2⁴ + 0×2³ + 0×2² + 0×2¹ + 0×2⁰
= 32 + 16 + 0 + 0 + 0 + 0
= 48
∴ 110000₂ = 48
(l) 110010₂
Solution:
110010₂
= 1×2⁵ + 1×2⁴ + 0×2³ + 0×2² + 1×2¹ + 0×2⁰
= 32 + 16 + 0 + 0 + 2 + 0
= 50
∴ 110010₂ = 50
(m) 110110₂
Solution:
110110₂
= 1×2⁵ + 1×2⁴ + 0×2³ + 1×2² + 1×2¹ + 0×2⁰
= 32 + 16 + 0 + 4 + 2 + 0
= 54
∴ 110110₂ = 54
(n) 111000₂
Solution:
111000₂
= 1×2⁵ + 1×2⁴ + 1×2³ + 0×2² + 0×2¹ + 0×2⁰
= 32 + 16 + 8 + 0 + 0 + 0
= 56
∴ 111000₂ = 56
(o) 111100₂
Solution:
111100₂
= 1×2⁵ + 1×2⁴ + 1×2³ + 1×2² + 0×2¹ + 0×2⁰
= 32 + 16 + 8 + 4 + 0 + 0
= 60
∴ 111100₂ = 60
(p) 111110₂
Solution:
111110₂
= 1×2⁵ + 1×2⁴ + 1×2³ + 1×2² + 1×2¹ + 0×2⁰
= 32 + 16 + 8 + 4 + 2 + 0
= 62
∴ 111110₂ = 62
6. Convert the following decimal numbers into binary number system:
(a) 15
Solution:
Therefore, 15 = 1111₂
(b) 18
Solution:
Therefore, 18 = 10010₂
(c) 21
Solution:
Therefore, 21 = 10101₂
(d) 24
Solution:
Therefore, 24 = 11000₂
(e) 26
Solution:
Therefore, 26 = 11010₂
(f) 30
Solution:
Therefore, 30 = 11110₂
(g) 33
Solution:
Therefore, 33 = 100001₂
(h) 36
Solution:
Therefore, 36 = 100100₂
(i) 39
Solution:
Therefore, 39 = 100111₂
(j) 41
Solution:
Therefore, 41 = 101001₂
(k) 43
Solution:
Therefore, 43 = 101011₂
(l) 46
Solution:
Therefore, 46 = 101110₂
(m) 49
Solution:
Therefore, 49 = 110001₂
(n) 52
Solution:
Therefore, 52 = 110100₂
(o) 64
Solution:
Therefore, 64 = 1000000₂
7. Find the value of A.
(a) For what value of A, the binary number A1101₂ has value 29?
Solution:
A1101₂ = A×24 + 1×23 + 1×22 + 0×21 + 1×20
= A×16 + 1×8 + 1×4 + 0×2 + 1×1
= 16A + 8 + 4 + 0 + 1
= 16A + 13
According to the question,
A1101₂ = 29
or, 16A + 13 = 29
or, 16A = 29 - 13
or, 16A = 16
or, A = 16/16 = 1
Therefore, A = 1
(b) For what value of A, the binary number 10A101₂ has value 45?
Solution:
10A101₂ = 1×25 + 0×24 + A×23 + 1×22 + 0×21 + 1×20
= 1×32 + 0×16 + A×8 + 1×4 + 0×2 + 1×1
= 32 + 0 + 8A + 4 + 0 + 1
= 8A + 37
According to the question,
10A101₂ = 45
or, 8A + 37 = 45
or, 8A = 45 - 37
or, 8A = 8
or, A = 8 / 8 = 1
Therefore, A = 1.
(c) For what value of A, the binary number 1A001₂ has value 25?
Solution:
1A001₂ = 1×24 + A×23 + 0×22 + 0×21 + 1×20
= 1×16 + A×8 + 0×4 + 0×2 + 1×1
= 16 + 8A + 0 + 0 + 1
= 17 + 8A
According to the question,
17 + 8A = 25
or, 8A = 25 - 17
or, 8A = 8
or, A = 8 / 8 = 1
Therefore, A = 1
(d) For what value of A, the binary number 110A11₂ has value 51?
Solution:
110A11₂ = 1×25 + 1×24 + 0×23 + A×22 + 1×21 + 1×20
= 1×32 + 1×16 + 0×8 + A×4 + 1×2 + 1×1
= 32 + 16 + 0 + 4A + 2 + 1
= 4A + 51
According to the question,
or, 4A = 51 - 51
or, 4A = 0
or, A = 0
Therefore, A = 0
8. If the lamp which is glowing on indicates 1 and the lamp which is glowing off indicates 0 of the binary system, write the binary numbers indicated by the following lamps and express them in decimal system.
(a) 110₂
Solution:
110₂ = 1×22 + 1×21 + 0×20
= 4 + 2 + 0
= 6
∴ 110₂ = 6
(b) 111₂
Solution:
111₂ = 1×22 + 1×21 + 1×20
= 4 + 2 + 1
= 7
∴ 111₂ = 7
(c) 100₂
Solution:
100₂ = 1×22 + 0×21 + 0×20
= 4 + 0 + 0
= 4
∴ 100₂ = 4
(d) 1100₂
Solution:
1100₂ = 1×23 + 1×22 + 0×21 + 0×20
= 8 + 4 + 0 + 0
= 12
∴ 1100₂ = 12
(e) 1110₂
Solution:
1110₂ = 1×23 + 1×22 + 1×21 + 0×20
= 8 + 4 + 2 + 0
= 14
∴ 1110₂ = 14
(f) 1111₂
Solution:
1111₂ = 1×23 + 1×22 + 1×21 + 1×20
= 8 + 4 + 2 + 1
= 15
∴ 1111₂ = 15
9. Convert the following decimal numbers into quinary numbers.
(a) 12
Solution:
Therefore, 12 = 22₅
(b) 15
Solution:
Therefore, 15 = 30₅
(c) 18
Solution:
Therefore, 18 = 33₅
(d) 20
Solution:
Therefore, 20 = 40₅
(e) 25
Solution:
Therefore, 25 = 100₅
(f) 30
Solution:
Therefore, 30 = 110₅
(g) 32
Solution:
Therefore, 32 = 112₅
(h) 44
Solution:
Therefore, 44 = 134₅
(i) 48
Solution:
Therefore, 48 = 143₅
(j) 50
Solution:
Therefore, 50 = 200₅
(k) 64
Solution:
Therefore, 64 = 224₅
(l) 72
Solution:
Therefore, 72 = 242₅
(m) 85
Solution:
Therefore, 85 = 320₅
(n) 90
Solution:
Therefore, 90 = 330₅
(o) 120
Solution:
Therefore, 120 = 440₅
(p) 140
Solution:
Therefore, 140 = 1030₅
(q) 150
Solution:
Therefore, 150 = 1100₅
Therefore, 150 = 1100₅
(r) 188
Solution:
Therefore, 188 = 1223₅
(s) 194
Solution:
Therefore, 194 = 1234₅
(t) 210
Solution:
Therefore, 210 = 1320₅
10. Convert the following quinary numbers into decimal numbers:
(a) 23₅
Solution:
23₅
= 2 × 5¹ + 3 × 5⁰
= 2 × 5 + 3 × 1 = 10 + 3 = 13
(b) 31₅
Solution:
31₅
= 3 × 5¹ + 1 × 5⁰
= 3 × 5 + 1 × 1 = 15 + 1 = 16
(c) 34₅
Solution:
34₅
= 3 × 5¹ + 4 × 5⁰
= 3 × 5 + 4 × 1 = 15 + 4 = 19
(d) 41₅
Solution:
41₅
= 4 × 5¹ + 1 × 5⁰
= 4 × 5 + 1 × 1 = 20 + 1 = 21
(e) 101₅
Solution:
101₅
= 1 × 5² + 0 × 5¹ + 1 × 5⁰
= 1 × 25 + 0 × 5 + 1 × 1 = 25 + 0 + 1 = 26
(f) 111₅
Solution:
111₅
= 1 × 5² + 1 × 5¹ + 1 × 5⁰
= 1 × 25 + 1 × 5 + 1 × 1 = 25 + 5 + 1 = 31
(g) 113₅
Solution:
113₅
= 1 × 5² + 1 × 5¹ + 3 × 5⁰
= 1 × 25 + 1 × 5 + 3 × 1 = 25 + 5 + 3 = 33
(h) 140₅
Solution:
140₅
= 1 × 5² + 4 × 5¹ + 0 × 5⁰
= 1 × 25 + 4 × 5 + 0 × 1 = 25 + 20 + 0 = 45
(i) 144₅
Solution:
144₅
= 1 × 5² + 4 × 5¹ + 4 × 5⁰
= 1 × 25 + 4 × 5 + 4 × 1 = 25 + 20 + 4 = 49
(j) 201₅
Solution:
201₅
= 2 × 5² + 0 × 5¹ + 1 × 5⁰
= 2 × 25 + 0 × 5 + 1 × 1 = 50 + 0 + 1 = 51
(k) 230₅
Solution:
230₅
= 2 × 5² + 3 × 5¹ + 0 × 5⁰
= 2 × 25 + 3 × 5 + 0 × 1 = 50 + 15 + 0 = 65
(l) 243₅
Solution:
243₅
= 2 × 5² + 4 × 5¹ + 3 × 5⁰
= 2 × 25 + 4 × 5 + 3 × 1 = 50 + 20 + 3 = 73
(m) 321₅
Solution:
321₅
= 3 × 5² + 2 × 5¹ + 1 × 5⁰
= 3 × 25 + 2 × 5 + 1 × 1 = 75 + 10 + 1 = 86
(n) 331₅
Solution:
331₅
= 3 × 5² + 3 × 5¹ + 1 × 5⁰
= 3 × 25 + 3 × 5 + 1 × 1 = 75 + 15 + 1 = 91
(o) 441₅
Solution:
441₅
= 4 × 5² + 4 × 5¹ + 1 × 5⁰
= 4 × 25 + 4 × 5 + 1 × 1 = 100 + 20 + 1 = 121
(p) 1031₅
Solution:
1031₅
= 1 × 5³ + 0 × 5² + 3 × 5¹ + 1 × 5⁰
= 1 × 125 + 0 × 25 + 3 × 5 + 1 × 1 = 125 + 0 + 15 + 1 = 141
(q) 1101₅
Solution:
1101₅
= 1 × 5³ + 1 × 5² + 0 × 5¹ + 1 × 5⁰
= 1 × 125 + 1 × 25 + 0 × 5 + 1 × 1 = 125 + 25 + 0 + 1 = 151
(r) 1224₅
Solution:
1224₅
= 1 × 5³ + 2 × 5² + 2 × 5¹ + 4 × 5⁰
= 1 × 125 + 2 × 25 + 2 × 5 + 4 × 1 = 125 + 50 + 10 + 4 = 189
(s) 1240₅
Solution:
1240₅
= 1 × 5³ + 2 × 5² + 4 × 5¹ + 0 × 5⁰
= 1 × 125 + 2 × 25 + 4 × 5 + 0 × 1 = 125 + 50 + 20 + 0 = 195
(t) 1321₅
Solution:
1321₅
= 1 × 5³ + 3 × 5² + 2 × 5¹ + 1 × 5⁰
= 1 × 125 + 3 × 25 + 2 × 5 + 1 × 1 = 125 + 75 + 10 + 1 = 211
11. Compare the numbers in different number systems:
(a) 100₅ and 11000₂
Solution:
100₅
= 1×52 + 0×51 + 0×50
= 25 + 0 + 0
= 25
Now,
11000₂
= 1×24 + 1×23 + 0×22 + 0×21 + 0×20
= 16 + 8 + 0 + 0 + 0
= 24
Therefore, 100₅ is greater than 11000₂ or 100₅ > 11000₂
(b) 101110₂ and 110₅
Solution:
101110₂
= 1×25 + 0×24 + 1×23 + 1×22 + 1×21 + 0×20
= 32 + 0 + 8 + 4 + 2 + 0
= 46
Now,
110₅
= 1×52 + 1×51 + 0×50
= 25 + 5 + 0
= 30
Therefore, 101110₂ is greater than 110₅ or 101110₂ > 110₅
(c) 1224₅ and 11001₂
Solution:
1224₅
= 1×53 + 2×52 + 2×51 + 4×50
= 125 + 50 + 10 + 4
= 189
Now,
11001₂
= 1×24 + 1×23 + 0×22 + 0×21 + 1×20
= 16 + 8 + 0 + 0 + 1
= 25
Therefore, 1224₅ is greater than 11001₂ or 1224₅ > 11001₂
(d) 101101₂ and 321₅
Solution:
101101₂
= 1×25 + 0×24 + 1×23 + 1×22 + 0×21 + 1×20
= 32 + 0 + 8 + 4 + 0 + 1
= 45
Now,
321₅
= 3×52 + 2×51 + 1×50
= 75 + 10 + 1
= 86
Therefore, 321₅ is greater than 101101₂ or 321₅ > 101101₂
(e) 1321₅ and 111110₂
Solution:
1321₅
= 1×53 + 3×52 + 2×51 + 1×50
= 125 + 75 + 10 + 1
= 211
Now,
111110₂
= 1×25 + 1×24 + 1×23 + 1×22 + 1×21 + 0×20
= 32 + 16 + 8 + 4 + 2 + 0
= 62
Therefore, 1321₅ is greater than 111110₂ or 1321₅ > 111110₂
(f) 101010₂ and 113₅
Solution:
101010₂
= 1×25 + 0×24 + 1×23 + 0×22 + 1×21 + 0×20
= 32 + 0 + 8 + 0 + 2 + 0
= 42
Now,
113₅
= 1×52 + 1×51 + 3×50
= 25 + 5 + 3
= 33
Therefore, 101010₂ is greater than 113₅ or 101010₂ > 113₅
12. Find the differences between the numbers:
(a) 100₅ and 11000₂
100₅
= 1×52 + 0×51 + 0×50
= 25 + 0 + 0
= 25
Now,
11000₂
= 1×24 + 1×23 + 0×22 + 0×21 + 0×20
= 16 + 8 + 0 + 0 + 0
= 24
Hence, the difference = 25 - 24 = 1
(b) 101110₂ and 110₅
101110₂
= 1×25 + 0×24 + 1×23 + 1×22 + 1×21 + 0×20
= 32 + 0 + 8 + 4 + 2 + 0
= 46
Now,
110₅
= 1×52 + 1×51 + 0×50
= 25 + 5 + 0
= 30
Hence, the difference = 46 - 30 = 16
(c) 1224₅ and 11001₂
1224₅
= 1×53 + 2×52 + 2×51 + 4×50
= 125 + 50 + 10 + 4
= 189
Now,
11001₂
= 1×24 + 1×23 + 0×22 + 0×21 + 1×20
= 16 + 8 + 0 + 0 + 1
= 25
Hence, the difference = 189 - 25 = 164
(d) 321₅ and 101101₂
321₅
= 3×52 + 2×51 + 1×50
= 75 + 10 + 1
= 86
Now,
101101₂
= 1×25 + 0×24 + 1×23 + 1×22 + 0×21 + 1×20
= 32 + 0 + 8 + 4 + 0 + 1
= 45
Hence, the difference = 86 - 45 = 41
(e) 1321₅ and 111110₂
1321₅
= 1×53 + 3×52 + 2×51 + 1×50
= 125 + 75 + 10 + 1
= 211
Now,
111110₂
= 1×25 + 1×24 + 1×23 + 1×22 + 1×21 + 0×20
= 32 + 16 + 8 + 4 + 2 + 0
= 62
Hence, the difference = 211 - 62 = 149
(f) 101010₂ and 113₅
101010₂
= 1×25 + 0×24 + 1×23 + 0×22 + 1×21 + 0×20
= 32 + 0 + 8 + 0 + 2 + 0
= 42
Now,
113₅
= 1×52 + 1×51 + 3×50
= 25 + 5 + 3
= 33
Hence, the difference = 42 - 33 = 9
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