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

Correct Option: (d) 200

The number of consecutive zeros at the end of a product is determined by the exponent of 10 in its prime factorization, which is the minimum of the exponents of 2 and 5. In the product P = 1^2 × 2^4 × 3^6 × ... × 25^50, the power of 2 is much higher than the power of 5. Thus, we only need to calculate the total exponent of 5. The terms in the product that contain 5 as a prime factor are: 5^10, 10^20, 15^30, 20^40, and 25^50. We can express these terms as: 5^10 = 5^10; 10^20 = (2 × 5)^20 = 2^20 × 5^20; 15^30 = (3 × 5)^30 = 3^30 × 5^30; 20^40 = (4 × 5)^40 = 2^80 × 5^40; 25^50 = (5^2)^50 = 5^100. Adding all the exponents of 5, we get: Total exponent of 5 = 10 + 20 + 30 + 40 + 100 = 200. Thus, there are exactly 200 consecutive zeros at the end of the integer. The correct option is (d).