Armstrong Number in Python
Check Whether a Given Number is an Armstrong Number or Not in Python
For a given integer the objective is to check whether or not the given integer is an Armstrong number or not. The Armstrong number is briefly defined in the section below.
Example
Input : 371
Output : It's an Armstrong Number
Check Whether a Given Number is an Armstrong Number or Not
Given an integer input, the objective is to check whether or not the given input variable is an Armstrong number. In order to do so, we check whether the sum of the digits of each number to the power the length of the number is equal to the number itself or not. If the number is the same as the original, it’s an Armstrong number. Mentioned below are a few of the Methods used to solve this problem,
- Method 1: Using Iteration
- Method 2: Using Recursion
We’ll see the working of the above mentioned methods in the upcoming sections. Please go through the below mentioned blue box for better understanding of the problem.
Example Input : 371 Output : It's an Armstrong Number Explanation : 371 can also be represented as 3^3 + 7^3 + 1^3 therefore it's an Armstrong Number.
Method 1: Using Iteration
In this method we’ll use for loop and while loop to check for Armstrong number.
Working
For an integer input number, we perform the following operations
- We break down the number into digits using “%” operator.
- We raise the digit to the power of the length of the number.
- Keep appending the value to sum variable.
- Check if the sum variable matches the number itself.
- If the above condition is satisfied, it’s an Armstrong Number.
Let’s implement the above logic in Python Language.
Python Code
number = 371
num = number
digit, sum = 0, 0
length = len(str(num))
for i in range(length):
digit = int(num%10)
num = num/10
sum += pow(digit,length)
if sum==number:
print("Armstrong")
else:
print("Not Armstrong")
Output
Armstrong
Method 2: Using Recursion
In this method we’ll use recursion to check for Armstrong number. To know more about recursion, check out Recursion in Python.
Working
For a given integer input, we do the following
- Define a recursive function checkArmstrong() with base case as num==0.
- Set the recursive step call as checkArmstrong(num/10, length, sum).
Let’s implement the above logic in Python Language.
Python Code
number = 371
num = number
sum =0
length = len(str(num))
def checkArmstrong(num,length,sum):
if num==0:
return sum
sum+=pow(int(num%10),length)
return checkArmstrong(num/10,length,sum)
if checkArmstrong(num,length,sum)==number:
print('Armstrong')
else:
print("Not Armstrong")
Output
Armstrong
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user_input = input(“Enter a number: “)
sum = 0
for _ in user_input:
sum += int((_))**len(user_input)
if sum == int(user_input):
print(“It’s a armstrong number”)
else:
print(“It’s not a armstrong number”)
num = int(input(“Enter a number: “))
# initialize sum
sum = 0
# find the sum of the cube of each digit
temp = num
while temp > 0:
digit = temp % 10
sum += digit ** 3
temp //= 10
# display the result
if num == sum:
print(num,”is an Armstrong number”)
else:
print(num,”is not an Armstrong number”)
n=input()
d=0
k=[i for i in n]
l=len(k)
for j in k:
p=int(j)
d=d+(p**l)
f=str(d)
h= [n for n in f]
if (h == k):
print(“Armstrong”)
else:
print(“not Armstrong”)
Find a Number is Armstrong or not:
n=int(input())
t=n
s=0
while n!=0:
r=n%10
s=s+(r**3)
n=n//10
print(s)
if t==s:
print(f”{s} is a armstrong number”)
else:
print(f”{s} not a armstrong number”)
x=input(‘Enter the number to be checked :’)
l=len(x)
s=0
for i in x:
s+= pow(int(i),l)
if (s==int(x)):
print(‘Armstrong !!!’)
else:
print(‘Not an Armstrong !!!’)
import math
k=int(input(‘num’))
l=[int(d) for d in str(k)]
s=0
while(k>0):
r=k%10
s=s+math.pow(r,len(l))
k=k//10
print(‘pal’,s)
q=int(input(“enter the num:”))
c=str(q)
s=[]
for i in c:
d=int(i)**3
s.append(d)
a=sum(s)
if c in str(a):
print(“yes it is a armstrong num”)
else:
print(“no it is not a armstrong number”)
Without importing libraries
num = int(input(‘enter: ‘))
num1 = str(num)
print(num1)
total_num = len(num1)
total = 0
for i in num1:
total = total + int(i)**total_num
print(total)
if num1 == str(total):
print(‘Armstrong number’)
else:
print(‘not armstrong number’)
another approach:
n = int(input(‘enter the no’))
result = 0
original = n
order = len(str(n))
while n > 0:
digits = n % 10
result += digits ** order
n = n // 10
if result == original:
print(‘no is armstrong’)
else:
print(‘no is not armstrong’)
num=int(input(“enter no:”))
temp=num
s=0
while num>0:
r=num%10
num=num//10
s=s+(r)**3
print(s)
if(s==temp):
print(“amstrong”)
else:
print(“not amstrong”)
x=int(input(‘enter the no :’))
r=0
u=len(str(x))
m=x
while(x!=0):
rem=x%10
r=r+pow(rem,u)
x=x//10
print(r)
if(r==m):
print(‘amstrong no’)
else:
print(‘not amstrong’)
def pow(g,a):
while(a!=0):
a=a-1
i=g*pow(g,a)
return i
By-gunjan soni
n=int(input())
k=len(str(n))
temp=n
sum=0
while(n!=0):
d=n%10
sum=sum+d**k
n=n//10
if(temp==sum):
print(“yes”)
else:
print(“no”)