Even if you were to assume that it would take 32 bits to represent a number like that (2 chars and an int), using the exact (well, ± 1) number of primes <= 10^23 (far less than what we're working with), it would still take (1925320391606818006727 * 32) /

/ 1024 / 1024 / 1024 / 1024 = (approximately) 7004274780 terabytes. Now, I do believe that 140 dvds is still far less than that, and that he's working with far more numbers. Just storing the numbers <= 10^23 would take about 1526037740000 dvds