Final answer:
The question involves calculating the number of trailing zeros in the product of factorials from 1! to 146!. The procedure involves counting the number of factors of 5 in the product, as there are more than enough factors of 2 to complement them. This is achieved by summing the counts of multiples of 5, 25, 125, etc., within the range of numbers from 1 to 146.
Step-by-step explanation:
The question asks us to calculate the number of consecutive zeros at the end of the decimal representation of the product (1!)(2!)(3!)...(146!). To solve this, we need to figure out how many times the number 10 (which is 2 multiplied by 5) is a factor in this product. Since every factorial from 5! onwards contains at least one factor of 5, and there are more factors of 2 than 5, we need to count the number of times 5 appears as a factor in this product.
To find the count of 5's factor, we divide the numbers by 5: 146/5 = 29, then by 25 (since 25 = 5² and contributes an extra factor of 5), and so on until the result is less than 1. We sum these counts to get the number of trailing zeros.
So, 146/5 + 146/25 + 146/125 gives us the total factors of 5. Then, each factorial contributes its count of 5s to the final product. We do not need to count 2's because there will always be more factors of 2 than 5, ensuring a 10 for each 5.