The number of zeros on the end of 599 factorial can be determined by factoring 599! The only part we are interested in are the factors of 2 and 5. Every time you can pair a 2 and 5 together, they multiply to ten and add a zero. Clearly there are more factors of 2 than 5. So five is the limiting case. The powers of five will determine the number of zeros.
Every fifth number 5 10, 15,...595 has a power of five
595/5 = 119
so there are 119 such numbers.
here for next division take take the first number which divisible by 5 and less than 119
115/5 = 23
20/5 = 4
So the number of zeros are,
119+23+4 = 146
there are 146 zeros