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 ?

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The correct answer is option D. 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.

Q9. 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 ?

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