RosettaCodeData/Task/Arbitrary-precision-integers-included-/Python/arbitrary-precision-integers-included-.py
2023-07-01 13:44:08 -04:00

3 lines
186 B
Python

>>> y = str( 5**4**3**2 )
>>> print ("5**4**3**2 = %s...%s and has %i digits" % (y[:20], y[-20:], len(y)))
5**4**3**2 = 62060698786608744707...92256259918212890625 and has 183231 digits