Python Program to check Abundant Number
Check Whether or Not the Number is an Abundant Number in Python
Given an integer input as the number, the objective is to check whether or the given integer number is an Abundant Number or not. Therefore, we’ll write a program to Check Whether or Not the Number is an Abundant Number in Python Language.
Example Input : 21 Output : It's not an Abundant Number.
Check Whether or Not the Number is an Abundant Number in Python Language
Given an integer input the objective is to check whether or not the given integer is an Abundant Number. To do so we’ll use loop to find the factors of the number except the number itself and sum them all up. We’ll check Whether the number is lower or greater than the sum. For a Number to be classified as Abundant, the sum of it’s factors must be greater than the number itself. Here are the method’s we’ll be using to Check Whether or Not the Number is an Abundant Number in Python Language,
- Method 1: Using Range until Number
- Method 2: Using Range until Sqrt( Number )
We’ll discuss the above mentioned methods in detail in the upcoming sections. Do check out the blue box given below for better understanding of the above mentioned questions.
Let's try and understand the concept better using an example
Example Input : Number = 12 Output : Yes, It's an Abundant Number Explanation : The Factors for the number 12 are, 1, 2, 3, 4 and 6. We don't want to include the number itself. Now the sum of the factors except the number itself is : 1 + 2 + 3 + 4 + 6 = 16 as the number 16>12 , the number itself. It's an abundant number.We'll discuss on how to check for it in the upcoming sections.
Method 1: Using Range until Number
Python Code
n = 12 sum=1 # 1 can divide any number for i in range(2,n): if(n%i==0): #if number is divisible by i add the number sum=sum+i if(sum>n): print(n,'is Abundant Number') else: print(n,'is not Abundant Number')
Output
12 is Abundant Number
Method 2: Using Range until Sqrt( Number )
Python Code
import math
n = 12
sum=1 # 1 can divide any number
i=2
while(i<=math.sqrt(n)):
if(n%i==0):
#if number is divisible by i add the number
if(n//i==i):
# if quotient is equal to divisor add only one of them
sum=sum+i
else:
sum=sum + i + n/i
i=i+1
if(sum>n):
print(n,"is Abundant Number")
else:
print(n,"is not Abundant Number")Output
12 is Abundant Number
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# Try This
def factors_sum(n):
s=0
for i in range(1,n//2+1):
if n%i==0:
s+=i
return s
n=int(input())
print(f”Factors Sum: {factors_sum(n)}”)
print(“Abundant Number” if factors_sum(n)>n else “Not Abundant Number”)
N = 12
ls = [i for i in range(1, N) if N % i == 0]
print(f”{N} is AbundantNumber” if sum(ls) > N else “Flase”)
a=[]
n=int(input())
for i in range(1,n):
if n%i==0 :
a.append(i)
print(a)
c=sum(a)
if(c>n):
print(‘abundant number’)
else:
print(‘not an abundant’)
n=int(input())
s=n
arr=[]
for i in range(1, n):
if n%i==0:
arr.append(i)
sum1=sum(arr)
if sum1>n:
print(‘It is an abundant number’)
else:
print(‘It is not an abundant number’)
# easy code
n=int(input(“Enter a number : “))
sum=0
for i in range(1,n):
if n%i==0:
sum=sum+i
print(“sum is”,sum)
print(“Input is ” , n)
if sum>n:
print(“Abandant Number”)
else:
print(“Not abandant number “)
def abundant(x):
y=[i for i in range(1,x) if(x%i==0)]
print(‘ abundant’ if(sum(y)>x) else ‘ not abundant’)
abundant (12)
x=int(input())
sum=0
for i in range(1,x):
if x%i==0:
sum=sum+i
if sum>x:
print(“yes”)
else:
print(“no”)
n=int(input())
sum=0
for i in range(1,n):
if(n%i==0):
sum=sum+i
if(sum>n):
print(‘Abundant Number’)
else:
print(‘Not Abundant Number’)
num=int(input(“Enter the number “))
sum=0
for i in range(1,num):
if num%i==0:
sum+=i
else:
continue
if sum>num:
print(num,”is an Abundant Number”)
else:
print(num,”is not an Abundant Number”)
n=int(input(“Enter the number:”))
sum=0
for i in range(1,n):
a=n%i
if(a==0):
sum=sum+i
if(sum>n):
print(“It is Abundant”)
else:
print(“It is Not”)