Task 1. Determine how many zeros the number $N!$ ends with!
Solution. Let $N=2014$. Among the first 2014 natural numbers, 402 numbers are divisible by 5, of which 80 numbers are divisible by 25. Among these 80 numbers, 16 numbers are divisible by 125, of which 3 numbers are divisible by 625. There are more even numbers among the first 2014 natural numbers than those divisible by 5. Therefore, the number of zeros at the end of the number 2014! is determined by the number of "fives" as divisors in this number. In total: $402+80+16+3=501$ zeros.

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