9. How many consecutive zeros are there at the end of the integer obtained in the product 1^2 × 2^4 × 3^6 × 4^8 ×…. × 25^50 ?

1. A. 50
2. B. 55
3. C. 100
4. D. 200

Answer: D

Explanation

The number of consecutive zeros at the end of a product is determined by the number of times 10 is a factor, which is equivalent to the number of pairs of (2 × 5) in its prime factorization. In this product, factors of 2 will be abundant. Therefore, we only need to count the factors of 5. The terms contributing factors of 5 are those whose base is a multiple of 5:
– 5^10 contributes 10 factors of 5.
– 10^20 = (2×5)^20 = 2^20 × 5^20 contributes 20 factors of 5.
– 15^30 = (3×5)^30 = 3^30 × 5^30 contributes 30 factors of 5.
– 20^40 = (4×5)^40 = 4^40 × 5^40 contributes 40 factors of 5.
– 25^50 = (5^2)^50 = 5^100 contributes 100 factors of 5.
Total number of factors of 5 = 10 + 20 + 30 + 40 + 100 = 200. Thus, there are 200 consecutive zeros at the end of the product. This question tests number theory concepts related to prime factorization and trailing zeros.