Understanding Trailing Zeros

Trailing zeros at the end of a number are formed by factors of 10. Since \(10 = 2 \times 5\), the number of trailing zeros in a product is determined by the number of pairs of prime factors 2 and 5 in its prime factorization.

In any product of positive integers, the number of factors of 2 is usually greater than or equal to the number of factors of 5. Therefore, the number of trailing zeros is typically limited by the number of factors of 5. However, it is always safest to count both the factors of 2 and the factors of 5 and take the minimum of the two counts.

Analyzing the Product: $1 \times 5 \times 10 \times \dots \times 60$

We need to find the total number of factors of 2 and the total number of factors of 5 in the prime factorization of the product:

\[ 1 \times 5 \times 10 \times 15 \times 20 \times 30 \times 35 \times 40 \times 45 \times 50 \times 55 \times 60 \] To do this, we will look at each number in the product and find its prime factors, specifically counting the factors of 5 and factors of 2.

Counting Factors of 5 in the Product

Let's list each number and determine how many times 5 appears as a prime factor:

Summing the number of factors of 5 from each number:

\(0 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 2 + 1 + 1 = 13\)

There are a total of 13 factors of 5 in the product.

Counting Factors of 2 in the Product

Now, let's list each number and determine how many times 2 appears as a prime factor:

Summing the number of factors of 2 from each number:

\(0 + 0 + 1 + 0 + 2 + 1 + 0 + 3 + 0 + 1 + 0 + 2 = 10\)

There are a total of 10 factors of 2 in the product.

Determining the Number of Trailing Zeros

The number of trailing zeros is the minimum of the total number of factors of 2 and 5.

• Total factors of 5 = 13
• Total factors of 2 = 10

Number of trailing zeros = \(\min(13, 10) = 10\).

Therefore, there are 10 zeroes at the end of the given product.

Revision Table: Counting Trailing Zeros

Additional Information: Prime Factors and Number Endings

Understanding prime factorization is key to solving problems involving the properties of numbers, such as counting trailing zeros. Every integer greater than 1 can be uniquely expressed as a product of prime numbers.

For trailing zeros, the primes 2 and 5 are critical because they multiply to give 10. A trailing zero is created every time a factor of 10 is formed in the product. We don't need to find the actual value of the massive product; we just need to count how many times the pair (2, 5) can be formed using the prime factors of all the numbers in the product.

For factorials (\(n!\)), the number of trailing zeros can be found efficiently using Legendre's formula, which counts the factors of 5 (as they are always fewer than factors of 2) by summing \(\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + \dots\). While our problem isn't a simple factorial, the principle of counting the minimum of factors of 2 and 5 remains the same.