Prove that the number of trailing zeros in n! is given by \{a/5\} + \{a/25\} + \{a/125\} +***,

The number of trailing zeros in \( n! \) (the factorial of \( n \)) is determined by the number of factors of 10 in \( n! \). Since \( 10 = 2 \times 5 \), each pair of a factor of 2 and a factor of 5 contributes to a trailing zero. However, in most factorials, there are more factors of 2 than factors of 5, so the number of trailing zeros is determined by the number of factors of 5.

### Step-by-Step Proof:

#### 1. Count the Factors of 5 in \( n! \)
The number of trailing zeros in \( n! \) corresponds to the number of factors of 5 in \( n! \).

- Consider \( n! = 1 \

Prove that the number of trailing zeros in n! is given by {a/5} + {a/25} + {a/125} + ***.

Explore how to prove the number of trailing zeros in n! by counting the factors of 5. This detailed step-by-step solution covers key concepts in number theory and factorials, focusing on divisibility rules. Suitable for students in Grades 10-12.

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