This assignment assesses your proficiency in converting an alphanumerical value
ID: 3833547 • Letter: T
Question
This assignment assesses your proficiency in converting an alphanumerical value among various base systems. You are to
complete the work and submit it on time for grading. Please show your work in order to receive partial credit should you
fail to answer the question correctly. Submit your work as an attachment using the Assignment link. Start early and email
your instructor for clarification on any ambiguity. I will be answering all email questions up to Friday by 5pm. Any email
questions arrive after 5pm on Friday may not be answered in time for you to work on your assignment and may cause you
to lose points. As the final assignment for this course, it means your course grade may be affected if you are unable to
have your question(s) answered in time. Start early!
In module 5, you viewed a video that introduces you to Binary (base 2), Octal (base 8), and Hexadecimal (base 16)
systems. The video also demonstrates how to convert between each base system and the nature number Decimal (base 10)
system, as well as converting among different base systems.
Today, your company received a letter from a healthcare provider stating that it will start using Base 32 system to
communicate numbers. The alphanumerical symbols it has chosen include numbers, from 0 to 9; and 22 letters, from A to
Z except I, L, O, and U. The following table shows each base 32 symbol and its corresponding Decimal value.
Base 32 symbol Decimal value
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
A 10
Base 32 symbol Decimal value
B 11
C 12
D 13
E 14
F 15
G 16
H 17
J 18
K 19
M 20
N 21
Base 32 symbol Decimal value
P 22
Q 23
R 24
S 25
T 26
Use the optional assignment description and replace each yellow highlighted box with the correct value to complete the conversions.
Base 32
P
R
0
V
E
Binary
B1
0
1
1
0
1
B2
0
0
0
0
0
B3
0
0
1
1
0
B4
1
0
1
1
1
B5
Octal
O1
3
O2
0
O3
1
O4
5
O5
Hexadecimal
H1
6
H2
0
H3
6
H4
Each correct answer in the Binary row worth 0.2 point.
B1 =
B2 =
B3 =
B4 =
B5 =
Each correct answer in the Octal row worth 0.4 point.
O1 =
O2 =
O3 =
O4 =
O5 =
Each correct answer in the Hexadecimal row worth 0.5 point.
H1 =
H2 =
H3 =
H4 =
Use the optional assignment description and replace each yellow highlighted box with the correct value to complete the conversions.
Base 32
P
R
0
V
E
Binary
B1
0
1
1
0
1
B2
0
0
0
0
0
B3
0
0
1
1
0
B4
1
0
1
1
1
B5
Octal
O1
3
O2
0
O3
1
O4
5
O5
Hexadecimal
H1
6
H2
0
H3
6
H4
Each correct answer in the Binary row worth 0.2 point.
B1 =
B2 =
B3 =
B4 =
B5 =
Each correct answer in the Octal row worth 0.4 point.
O1 =
O2 =
O3 =
O4 =
O5 =
Each correct answer in the Hexadecimal row worth 0.5 point.
H1 =
H2 =
H3 =
H4 =
Base 32
P
R
0
V
E
Binary
B1
0
1
1
0
1
B2
0
0
0
0
0
B3
0
0
1
1
0
B4
1
0
1
1
1
B5
Octal
O1
3
O2
0
O3
1
O4
5
O5
Hexadecimal
H1
6
H2
0
H3
6
H4
Explanation / Answer
Since we know Base32 to Decimal mapping, converting Base32 to Binary is simply same as converting the corresponding Decimal value.
P (Base32) is 22 (Decimal), hence its 10110 (Binary) similarly others. Lets put that in a table as follows
Base32 Decimal Binary
P 22 10110
R 24 11000
0 0 00000
V 27 11011
E 14 01110
We can now answer B1, B2, B3 and B4 from the above table as
B1=1
B2=1
B3=0
B4=1
B5=0
Now 3rd box Octal is from Binary since we can see that 3 bits of Binary are grouped to form Octal numbers in the next row.
So O1 corresponds to B1 and O2 to 01B2 and so on as shown in the given table. Lets put them in the table again
Octal Corresponding Binary
O1 B1 or (00B1 by padding with zeros)
O2 01B2
O3 00B3
O4 10B4
O5 11B5
Since we know B1-B5 values lets replace them and find O1-O5 values
Octal Corresponding Binary
O1 1 (or 001)
O2 011
O3 000
O4 101
O5 110
So the O1-O5 values are
O1=1
O2=3
O3=0
O4=5
O5=6
Similarly 2 Binary bits are grouped to get Hexadecimal values as shown in the given table. So we get,
Hexadecimal Corresponding Binary
H1 B1
H2 1B200
H3 0011
H4 111B5
Replacing B values we get
Hexadecimal Corresponding Binary
H1 1
H2 1100
H3 0011
H4 1110
So the values are
H1=1
H2=12
H3=3
H4=14