Course Content

Numeral Systems 101

1. Binary Numeral System

Welcome on Board!Get Acquainted with Binary Code Decipher Binary Number Decipher List of Binary Numbers Cipher Some Numbers Hone your Ciphering Skills Dates in Binary

2. Octal Numeral system

Get Acquainted with the Octal Numeral System Practice Ciphering

3. Hexadecimal Numeral system

Deciphering Challenge One More Ciphering Challenge

4. Revelation

Write the Code in One Line Harder Converting Converting Using Format Method

Practice Ciphering

So far so good 😉.

As you already know how to convert decimal numbers to binary, a useful step is to learn how to do the opposite action. As you get familiar with the previous steps this one is not going to be a challenge for you. 1 in binary code is still one, 0 is still zero, but here you have 8 numbers so 9 should be converted properly 0->0 1->1 2->2 3->3 4->4 5->5 6->6 7->7.In the example, we define 1234 as a decimal number.

You need to divide the number by 8, and take down the remainder of the division.
Then you should calculate the received number and implement the first step to it.
You can stop if division results in 0.
Rewrite remainders in the reversed order.

As mentioned, the more practice the better. Here you are going to make a complex task where you will work not only with a list of converted numbers but with string: it means that you should not only append remainders but convert it to strings firstly. Try to do the task to see beautiful representations in string.


              1234567891011121314151617181920212223242526
            
# Defininig the decimal number
decimal_number = 4321
# Creating a list for storing the converted binary number
octal_number = [ ]
# The text should be realised here due to the reason that further the decimal number will be changed
print("The number in decimal numeral system is:", decimal_number)
# Check if the decimal number is zero: further we will convert it till the number is zero
if decimal_number == 0:
    # 0 in the octal numeral system is the representation of 0 in decimal one, so we append 0 to the list of the octal numbers
    octal_number.append(str(0))
#otherwise
else:
    # The loop executes till the number is zero
    while decimal_number != 0:
        # Counting the remainder of dividing by eight
        remainder = decimal_number % 8
        # Appending the converted remainder for creating an octal representation
        octal_number.append(str(remainder))
        # This operation allows to decrease number eight times and work with the integer part of new one
        decimal_number = decimal_number//8
    # Reversing the string
    octal_number = octal_number[::-1]
# Joining all items of the string to make it more readable
octal_number = " ".join(octal_number)
# Printing the result
print("The number in octal numeral system is:", octal_number)

Task

Write the code that will decode the number 3196 from the decimal numeral system to octal. Fill the gaps and follow the algorithm. If everything is correct you will receive another special number 🤯, the explanation is waiting for you at the end of the chapter.

Define an empty list for storing the octal_number.
Print the decimal_number variable.
Check if the decimal_number variable is 0.
Append 0 if decimal_number variable is 0.
Define the loop which works while decimal number is not 0.
Count the remainder of division decimal_number by 8.
Append converted remainder to the list of octal numbers.
Decrese decimal_number using integer division it by 8.
Make the string of octal numbers reversed.

Note
I bet you've received 6174. This is a Kaprekar's and it is called after the mathematician D.R. Kaprekar.Try to do the simple calculation with a number that consists of at least two different digits: 5678 for instance. Firstly write this number in descending and ascending order and subtract it: 8765-5678 = 3087, then we will do the same, but with the number 3087 8730 - 0378 = 8352, the same here 8532 - 2358 = 6174 7641 - 1467 = 6774. You can apply the same algorithm to different numbers and you will receive 6174 every time.

Switch to desktop for real-world practiceContinue from where you are using one of the options below

Everything was clear?

Thanks for your feedback!

Section 2. Chapter 2

Practice Ciphering

So far so good 😉.

You need to divide the number by 8, and take down the remainder of the division.
Then you should calculate the received number and implement the first step to it.
You can stop if division results in 0.
Rewrite remainders in the reversed order.


              1234567891011121314151617181920212223242526
            
# Defininig the decimal number
decimal_number = 4321
# Creating a list for storing the converted binary number
octal_number = [ ]
# The text should be realised here due to the reason that further the decimal number will be changed
print("The number in decimal numeral system is:", decimal_number)
# Check if the decimal number is zero: further we will convert it till the number is zero
if decimal_number == 0:
    # 0 in the octal numeral system is the representation of 0 in decimal one, so we append 0 to the list of the octal numbers
    octal_number.append(str(0))
#otherwise
else:
    # The loop executes till the number is zero
    while decimal_number != 0:
        # Counting the remainder of dividing by eight
        remainder = decimal_number % 8
        # Appending the converted remainder for creating an octal representation
        octal_number.append(str(remainder))
        # This operation allows to decrease number eight times and work with the integer part of new one
        decimal_number = decimal_number//8
    # Reversing the string
    octal_number = octal_number[::-1]
# Joining all items of the string to make it more readable
octal_number = " ".join(octal_number)
# Printing the result
print("The number in octal numeral system is:", octal_number)

Task

Define an empty list for storing the octal_number.
Print the decimal_number variable.
Check if the decimal_number variable is 0.
Append 0 if decimal_number variable is 0.
Define the loop which works while decimal number is not 0.
Count the remainder of division decimal_number by 8.
Append converted remainder to the list of octal numbers.
Decrese decimal_number using integer division it by 8.
Make the string of octal numbers reversed.

Note
I bet you've received 6174. This is a Kaprekar's and it is called after the mathematician D.R. Kaprekar.Try to do the simple calculation with a number that consists of at least two different digits: 5678 for instance. Firstly write this number in descending and ascending order and subtract it: 8765-5678 = 3087, then we will do the same, but with the number 3087 8730 - 0378 = 8352, the same here 8532 - 2358 = 6174 7641 - 1467 = 6774. You can apply the same algorithm to different numbers and you will receive 6174 every time.

Switch to desktop for real-world practiceContinue from where you are using one of the options below

Everything was clear?

Thanks for your feedback!

Section 2. Chapter 2

Practice Ciphering

So far so good 😉.

You need to divide the number by 8, and take down the remainder of the division.
Then you should calculate the received number and implement the first step to it.
You can stop if division results in 0.
Rewrite remainders in the reversed order.


              1234567891011121314151617181920212223242526
            
# Defininig the decimal number
decimal_number = 4321
# Creating a list for storing the converted binary number
octal_number = [ ]
# The text should be realised here due to the reason that further the decimal number will be changed
print("The number in decimal numeral system is:", decimal_number)
# Check if the decimal number is zero: further we will convert it till the number is zero
if decimal_number == 0:
    # 0 in the octal numeral system is the representation of 0 in decimal one, so we append 0 to the list of the octal numbers
    octal_number.append(str(0))
#otherwise
else:
    # The loop executes till the number is zero
    while decimal_number != 0:
        # Counting the remainder of dividing by eight
        remainder = decimal_number % 8
        # Appending the converted remainder for creating an octal representation
        octal_number.append(str(remainder))
        # This operation allows to decrease number eight times and work with the integer part of new one
        decimal_number = decimal_number//8
    # Reversing the string
    octal_number = octal_number[::-1]
# Joining all items of the string to make it more readable
octal_number = " ".join(octal_number)
# Printing the result
print("The number in octal numeral system is:", octal_number)

Task

Define an empty list for storing the octal_number.
Print the decimal_number variable.
Check if the decimal_number variable is 0.
Append 0 if decimal_number variable is 0.
Define the loop which works while decimal number is not 0.
Count the remainder of division decimal_number by 8.
Append converted remainder to the list of octal numbers.
Decrese decimal_number using integer division it by 8.
Make the string of octal numbers reversed.

Note
I bet you've received 6174. This is a Kaprekar's and it is called after the mathematician D.R. Kaprekar.Try to do the simple calculation with a number that consists of at least two different digits: 5678 for instance. Firstly write this number in descending and ascending order and subtract it: 8765-5678 = 3087, then we will do the same, but with the number 3087 8730 - 0378 = 8352, the same here 8532 - 2358 = 6174 7641 - 1467 = 6774. You can apply the same algorithm to different numbers and you will receive 6174 every time.

Switch to desktop for real-world practiceContinue from where you are using one of the options below

Everything was clear?

Thanks for your feedback!

So far so good 😉.

You need to divide the number by 8, and take down the remainder of the division.
Then you should calculate the received number and implement the first step to it.
You can stop if division results in 0.
Rewrite remainders in the reversed order.


              1234567891011121314151617181920212223242526
            
# Defininig the decimal number
decimal_number = 4321
# Creating a list for storing the converted binary number
octal_number = [ ]
# The text should be realised here due to the reason that further the decimal number will be changed
print("The number in decimal numeral system is:", decimal_number)
# Check if the decimal number is zero: further we will convert it till the number is zero
if decimal_number == 0:
    # 0 in the octal numeral system is the representation of 0 in decimal one, so we append 0 to the list of the octal numbers
    octal_number.append(str(0))
#otherwise
else:
    # The loop executes till the number is zero
    while decimal_number != 0:
        # Counting the remainder of dividing by eight
        remainder = decimal_number % 8
        # Appending the converted remainder for creating an octal representation
        octal_number.append(str(remainder))
        # This operation allows to decrease number eight times and work with the integer part of new one
        decimal_number = decimal_number//8
    # Reversing the string
    octal_number = octal_number[::-1]
# Joining all items of the string to make it more readable
octal_number = " ".join(octal_number)
# Printing the result
print("The number in octal numeral system is:", octal_number)

Task

Define an empty list for storing the octal_number.
Print the decimal_number variable.
Check if the decimal_number variable is 0.
Append 0 if decimal_number variable is 0.
Define the loop which works while decimal number is not 0.
Count the remainder of division decimal_number by 8.
Append converted remainder to the list of octal numbers.
Decrese decimal_number using integer division it by 8.
Make the string of octal numbers reversed.

Note
I bet you've received 6174. This is a Kaprekar's and it is called after the mathematician D.R. Kaprekar.Try to do the simple calculation with a number that consists of at least two different digits: 5678 for instance. Firstly write this number in descending and ascending order and subtract it: 8765-5678 = 3087, then we will do the same, but with the number 3087 8730 - 0378 = 8352, the same here 8532 - 2358 = 6174 7641 - 1467 = 6774. You can apply the same algorithm to different numbers and you will receive 6174 every time.

Switch to desktop for real-world practiceContinue from where you are using one of the options below

Section 2. Chapter 2

Switch to desktop for real-world practiceContinue from where you are using one of the options below