017 - sum a power of 2

See Problem 17.

Great artists steal. The prodigious Andy Dustman has a function for spoken number which I've cannibalised below, with a few test calls:

numbers = {
         0: 'zero',
         1: 'one',
         2: 'two',
         3: 'three',
         4: 'four',
         5: 'five',
         6: 'six',
         7: 'seven',
         8: 'eight',
         9: 'nine',
         10: 'ten',
         11: 'eleven',
         12: 'twelve',
         13: 'thirteen',
         14: 'fourteen',
         15: 'fifteen',
         16: 'sixteen',
         17: 'seventeen',
         18: 'eighteen',
         19: 'nineteen',
         20: 'twenty',
         30: 'thirty',
         40: 'forty',
         50: 'fifty',
         60: 'sixty',
         70: 'seventy',
         80: 'eighty',
         90: 'ninety'
}

def number_as_word (num):
        if num in numbers:
            return numbers[num]
        if (num < 100):
            return number_as_word ((num/10)*10) + '-' + number_as_word (num%10)
        elif (num < 1000):
            stem = number_as_word (num/100) + ' ' + 'hundred'
            remainder = num % 100
        elif (num < 1000000):
            stem = number_as_word (num/1000) + ' ' + 'thousand'
            remainder = num % 1000
        if (remainder):
                stem += ' and ' + number_as_word (remainder)
        return stem

print number_as_word (7)
print number_as_word (17)
print number_as_word (81)
print number_as_word (127)
print number_as_word (900)
print number_as_word (901)
print number_as_word (1000)

Then just iterate over all the numbers, and count their length (after removing spaces and hyphens):

total = 0;
for x in xrange (1, 1001):
        number = number_as_word (x)
        total += len (number.replace (' ', '').replace ('-', ''))
print total

The answer is 21124. Computation time is 0.005671s.