Embedly problem 3 in python

embedly3.py

import math
totalWords = 0
for denominator in range(1, 901):
        numOccurrences = 2520.0 / denominator
        numOccurrences = math.floor(numOccurrences + 0.5)
        totalWords = totalWords + numOccurrences
print 'the text has %d total words' % totalWords
halfTheWords = totalWords / 2
for denominator in range(1, 901):
        halfTheWords -= 2520 / denominator
        print "round %d: %d" % ( denominator, halfTheWords )
        if halfTheWords < 0:
                print denominator
                break