In [ ]:

    

'''
Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:

1634 = 1^4 + 6^4 + 3^4 + 4^4
8208 = 8^4 + 2^4 + 0^4 + 8^4
9474 = 9^4 + 4^4 + 7^4 + 4^4
As 1 = 1^4 is not a sum it is not included.

The sum of these numbers is 1634 + 8208 + 9474 = 19316.

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
'''


    

In [8]:

    

# find bounds: for example, smallest 5 digit number: 10000
# fifth power
# sum is 1^5 = 1 
# maximum sum is 5 * 9**5
# 5 * 9**5
# 295245
# 6 digits: 100000 
# 7 digits: 1 000 000 = 10**(digits - 1)
# 7 * 9**5 # (digits) * 9**(power)
# 413343
# so no seven digit number can be expressed as a sum of the fifth powers of its digits, because the highest possible sum 
# for 9999999 is less than 7 digits 

# so lets define a boundary first 

def findNumsAsPowers(power):
    '''Returns a set of numbers which can be expressed as the sum of the power-th power of their digits.'''
    
    nums = set()
    i = 1 # i is the number of digits in our number, which we have to put a limit on
    
    
    while ((10**(i-1)) < (i * (9**power))):
        i += 1
    
    # now use i as a limit 
    
    for j in range(2, (10**(i-1))):
        ints = [int(k) for k in str(j)] # split the number into digits
        powersum = sum((l**power) for l in ints) # calculate power sum
        if powersum == j: # save if digit power sum equals the number 
            nums.add(j)
        
    return(nums)


    

In [11]:

    

sum(findNumsAsPowers(5))


    

    
Out[11]:





443839