Math Problem Statement
Prove that the number of trailing zeros in n! is given by {a/5} + {a/25} + {a/125} +***,
Solution
The number of trailing zeros in (the factorial of ) is determined by the number of factors of 10 in . Since , 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
The number of trailing zeros in corresponds to the number of factors of 5 in .
- Consider ( n! = 1 \
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Math Problem Analysis
Mathematical Concepts
Factorial
Divisibility
Number Theory
Formulas
Number of trailing zeros in n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ...
Theorems
Fundamental Theorem of Arithmetic
Divisibility Rules
Suitable Grade Level
Grades 10-12