[Tim Peters] > You would in this case, and that would be wrong. In fp you'd get an > approximation to the exact n * (1./5 + 1./5**2 + ...) == n/4. (use > the rule for the sum of an infinite geometric series). For example, > that way you'd compute that 4! == 24 has 4/4 == 1 trailing zero, > instead of the correct 4 // 5 == 0 trailing zeroes, and that 9! == > 362880 has 9/4 == 2.25 trailing zeroes instead of the correct 9 // 5 > == 1 trailing zero.
well ... you're right, of course. > Nope again. Count the number of trailing zeros in 100! more carefully. since i'm not able to count further than to 10 past midnight, i used echo -n "000000000000000000000000" | wc -c to do the work. and because i'm a confused person i forgot the "-n" :-) good night, christian _______________________________________________ Tutor maillist - [email protected] http://mail.python.org/mailman/listinfo/tutor
