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 n!n! (the factorial of nn) is determined by the number of factors of 10 in n!n!. Since 10=2×510 = 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!n!

The number of trailing zeros in n!n! corresponds to the number of factors of 5 in n!n!.

  • Consider ( n! = 1 \

Ask a new question for Free

By Image

Drop file here or Click Here to upload

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