Finding Consecutive Zeros at the End of the Product 1! * 2! * 3! * ... * 100!

In the realm of combinatorial mathematics and number theory, a common challenge is to determine the number of consecutive zeros at the end of the product of factorials. In this article, we will explore a specific instance: the product of the factorials from 1! to 100!.

Introduction to the Problem

For the product of the factorials of the first 100 positive integers, ( 1! times 2! times 3! times ldots times 100! ), we seek to determine the number of consecutive zeros at the end of this product. This is closely related to how many times the product can be divided by 10, as each pair of factors 2 and 5 results in a trailing zero. However, since factorials contain a higher frequency of the factor 2, we focus on counting the number of times the factor 5 appears.

Counting the Factors of 5

To solve this, we use the formula for the number of times a factor ( p ) appears in ( n! ). Specifically, for factor 5:

[ sum_{k1}^{infty} leftlfloor frac{n}{5^k} rightrfloor ]

We need to calculate the number of factors of 5 for each factorial from 1! to 100! and sum these counts.

Step-by-Step Calculation

Step 1: Count the Factors of 5 for Each ( n! )

Let's calculate the number of factors of 5 for each ( n! ) from 1 to 100:

For ( 1! ) to ( 4! ): 0 For ( 5! ): 1 For ( 6! ) to ( 9! ): 1 For ( 10! ): 2 For ( 11! ) to ( 14! ): 2 For ( 15! ): 3 Continuing in this pattern, we can find the factors of 5 for ( n! ) up to 100!

Step 2: General Calculation for ( n! )

We can summarize the general calculation for ( n! ).

If ( n 5k ), the number of factors of 5 is ( k ). If ( n 5k j ), where ( j 1, 2, 3, 4 ), the number of factors of 5 remains ( k ) because adding 1, 2, 3, or 4 does not add additional factors of 5.

Step 3: Sum the Factors for ( 1! times 2! times ldots times 100! )

We now sum the counts of factors of 5 for all the factorials from 1 to 100:

[ sum_{n1}^{100} leftlfloor frac{n}{5} rightrfloor sum_{n1}^{100} leftlfloor frac{n}{25} rightrfloor ldots ]

This sum can be broken down:

For multiples of 5 up to 100, there are 20 such numbers: ( 5, 10, 15, ldots, 100 ). For multiples of 25 up to 100, there are 4 such numbers: ( 25, 50, 75, 100 ).

Therefore, the total number of factors of 5 in ( 1! times 2! times ldots times 100! ) is:

[ 20 4 24 ]

Conclusion

Thus, the number of consecutive zeros at the end of the product ( 1! times 2! times 3! times ldots times 100! ) is 24.

This problem showcases the application of combinatorial mathematics and divisibility rules in number theory, providing a practical insight into factorial products and their properties.