Python: Perfect Number
Python: Non-abundant sums
Problem No.23 @ ProjectEuler
In projecteuler Problem No. 23 talking about Non-abundant sums to clear this it state three categories for a numbers.
1. A number N called Perfect Number if the sum of its divisors are equal to the number it self. So 28 is a perfect number because the divisors of 28 are 1,2,4,7,14 they equal to 28. ( 1 + 2 + 4 + 7 + 1 4 = 28)
2. A number N is called Deficient if the sum of its proper divisors is less than the number it self N.
3. And a number N is called Abundant if this sum of its proper divisors is exceeds N (number it self)
Enhancement: So instead of solving problem No.23, we will write a code to determent the group that a number belongs to. As we trying to make a general cods, we will ask the user to enter a number then we will call the function to collect all the divisors of N and get its sum and gives it a name according to the classification we just talk about.
The Code:
# Python: Non-abundant sums
# Problem No.23 @ ProjectEuler
# We solve it in my way
num_divisors=[]
def get_num_group (num):
for x in range(1,num):
if num%x == 0:
num_divisors.append(x)
num =int(input(‘Enter a number: ‘))
div_sum =0
get_num_group(num)
for each in num_divisors:
div_sum = div_sum + each
print(‘\n Divisors of {} are,’.format(num),num_divisors)
print(‘ The sum of the Divisors is ‘,div_sum)
if div_sum == num :
print (‘ The number {} is Perfect Number’.format(num))
elif div_sum < num :
print (‘ The number {} is Deficient ‘.format(num))
else:
print (‘ The number {} is Abundant ‘.format(num))
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