Correct Option

The number of consecutive zeros at the end of an integer is determined by the lowest power of 10 in its prime factorization. Since 10 = 2 × 5, this is equivalent to finding the minimum count between the prime factors 2 and 5. In products of this nature, the count of factor 2 is typically much higher than the count of factor 5. Therefore, the number of zeros is limited by the total count of the prime factor 5.

The given product is $1^{2} \times 2^{4} \times 3^{6} \times 4^{8} \times . . . \times 25^{50}$ . We need to identify terms that contribute factors of 5:

From $5^{10}$ : This term contributes $5^{10}$ to the product.
From $10^{20}$ : This term is $(2 \times 5)^{20} = 2^{20} \times 5^{20}$ , contributing $5^{20}$ .
From $15^{30}$ : This term is $(3 \times 5)^{30} = 3^{30} \times 5^{30}$ , contributing $5^{30}$ .
From $20^{40}$ : This term is $(4 \times 5)^{40} = (2^{2} \times 5)^{40} = 2^{80} \times 5^{40}$ , contributing $5^{40}$ .
From $25^{50}$ : This term is $(5^{2})^{50} = 5^{100}$ , contributing $5^{100}$ .

The total power of 5 in the product is the sum of the exponents from these terms:

$10 + 20 + 30 + 40 + 100 = 200$

Thus, the product contains $5^{200}$ as a factor. As the number of factors of 2 will be significantly higher than 200 in this product, the number of consecutive zeros at the end of the integer is 200.

Incorrect Options

Options (1) 50, (2) 55, and (3) 100 are incorrect. These values do not represent the cumulative contribution of the prime factor 5 from all relevant terms in the product. The detailed calculation, considering all multiples of 5 up to 25 and their respective exponents, yields a total power of 5 as 200.