Python Program to Check Armstrong Number

In this article, we've created some programs in Python, to check whether a number entered by user is an Armstrong number or not. Here are the list of programs:

• Simple Program to Check Armstrong Number
• Check Armstrong Number using user-defined Function
• Using class

But before starting these programs, let me clear the thing about Armstrong number, that what it is.

What is an Armstrong Number ?

A number that equals to the sum of its own digits, where each digit raised to the power of number of digits. For example, 153 is an Armstrong number, because:

153 = 13 + 53 + 33
    = 1 + 125 + 27
    = 153

The result (153) is equal to the number (153) itself. So it is an Armstrong number.

Note - Because the total number of digit in 153 is 3, so each of its digit raised to the power of 3.

Note - 1, 2, 3, 4, 5, 6, 7, 8, and 9 are all Armstrong numbers.

Check Armstrong Number

To check whether a given number is an Armstrong number or not in Python, you have to ask from user to enter a number, then apply the formula, check and print the message as shown in the program given below:

print("Enter the Number: ")
num = int(input())

temp = num
noOfDigit = 0
res = 0
while num>0:
    num = int(num/10)
    noOfDigit = noOfDigit+1
num = temp
while num>0:
    rem = num%10
    pow = 1
    i = 0
    while i<noOfDigit:
        pow = pow*rem
        i = i+1
    res = res+pow
    num = int(num/10)

if res==temp:
    print("\nThe number is an Armstrong Number")
else:
    print("\nThe number is not an Armstrong Number")

Here is the initial output produced by this Python program:

Now supply the input say 1634 and press ENTER key to check whether it is an armstrong number or not. Here is its sample run with this user input:

Since 1634 is a four-digit number, therefore:

1634 = 14 + 64 + 34 + 44
     = 1 + 1296 + 81 + 256
     = 1297 + 337
     = 1634

In above program, the following block of code:

while num>0:
num = int(num/10)
noOfDigit = noOfDigit+1

is used to count the total number of digits of given number. The dry run of this block of code with user input 1634 goes like:

• Initial value, noOfDigit=0
• The condition, num>0 or 1634>0 evaluates to be true, therefore program flow goes to the body of while loop and executes both the statements
• That is, int(num/10) or int(1634/10) or 163 initialized to num. So num=163 now. And noOfDigit+1 or 0+1 or 1 gets initialized to noOfDigit. So noOfDigit=1
• Now the condition of while loop again gets evaluated, that is the condition num>0 or 163>0 evaluates to be true again, therefore program flow goes inside its body and executes that two statements again. This process continues, until the condition evaluates to be false
• In this way, after exiting from this loop, the variable noOfDigit holds its value as 4

And the following block of code:

while i<noOfDigit:
    pow = pow*rem
    i = i+1

is used to calculate the value of digit's raised to the power of number of digit. The dry run of this block of code goes like:

• Initial values, i=0, noOfDigit=4, pow=1, rem=4 (because 1634%10 is equal to 4)
• Now the condition i<noOfDigit or 0<4 evaluates to be true, therefore program flow goes inside its body
• And pow*rem or 1*4 or 4 gets initialized to pow. So pow=4, and the value of i gets incremented by 1. So i=1
• Since the condition at second evaluation, that is i<noOfDigit or 1<4 evaluates to be true again, therefore program flow again goes inside the loop and pow*rem or 4*4 or 16 initialized to pow. So pow=16 and the value of i=2
• At third time also, the condition i<noOfDigit or 2<4 evaluates to be true, therefore program flow again goes inside the loop. This process continues until the condition evaluates to be true
• In this way, after exiting from this loop, the variable pow holds its value as 1*4*4*4*4 or 256 that is equal to 4⁴, where 4 is the digit (last digit of number) and power 4 indicates to total number of digit

After second evaluation of this block of code, pow holds its value as 3*3*3*3, at third time pow holds 6*6*6*6 and at fourth time pow holds 1*1*1*1

Modified Version of Previous Program

This is the modified version of previous program. In this program, we've used end to skip printing of an automatic newline using print(). And rem**noOfDigit is used to find rem^noOfDigit value. The str() method is used to convert any type of value to string type value. Rest of the things are similar to previous program.

print("Enter the Number: ", end="")
num = int(input())

temp = num
noOfDigit = len(str(num))
res = 0
while num>0:
    rem = num%10
    res = res + (rem ** noOfDigit)
    num = int(num/10)

if res==temp:
    print("\n" +str(temp)+ " is an Armstrong Number")
else:
    print("\n" +str(temp)+ " is not an Armstrong Number")

Here is its sample run with user input, 9:

Check Armstrong Number using Function

This program uses used-defined function named checkArmstrongNum() to check whether a number entered by user at run-time is an Armstrong number or not.

def checkArmstrongNum(x):
    noOfDigit = 0
    res = 0
    temp = x
    while x>0:
        x = int(x/10)
        noOfDigit = noOfDigit + 1
    x = temp
    while x>0:
        rem = x%10
        pow = 1
        i = 0
        while i<noOfDigit:
            pow = pow * rem
            i = i + 1
        res = res + pow
        x = int(x/10)

    if res==temp:
        return 1
    else:
        return 0

print("Enter the Number: ", end="")
num = int(input())

chk = checkArmstrongNum(num)
if chk==1:
    print(num, "is an Armstrong Number")
else:
    print(num, "is not an Armstrong Number")

Here is its sample run with user input, 371:

Here is another sample run with user input, 532:

Check Armstrong Number using Class

This is the last program of this article on checking Armstrong number, using class, an object-oriented feature of Python:

class CodesCracker:
    def checkArmstrongNum(self, x):
        noOfDigit = len(str(x))
        res = 0
        temp = x
        while x>0:
            rem = x%10
            res = res + (rem ** noOfDigit)
            x = int(x/10)
        if res==temp:
            return 1
        else:
            return 0

print("Enter the Number: ", end="")
num = int(input())

obj = CodesCracker()
chk = obj.checkArmstrongNum(num)

if chk==1:
    print(num, "is an Armstrong Number")
else:
    print(num, "is not an Armstrong Number")

An object obj is created of type CodesCracker class to access its member function named checkArmstrongNum() using dot (.) operator.

Same Program in Other Languages

• Java Check Armstrong or Not
• C Check Armstrong or Not
• C++ Check Armstrong or Not

Python Online Test