Q. How many consecutive zeros are there at the end of the integer obtained in the product 1² × 2⁴ × 3⁶ × 4⁸ × … × 25⁵⁰?

Explanation : 

Calculating Consecutive zeros

To determine the number of consecutive zeros at the end of the integer obtained from the product 1^2 × 2^4 × 3^6 × 4^8 × … × 25^50, we need to find the highest power of 10 that divides this product. Since 10 = 2 × 5, we need to count the number of factors of 2 and 5 in the product and take the minimum of these counts.

First, let’s express the product in a more general form:
Product from k=1 to 25 of k^(2k)

Counting the Factors of 5:
We need to count the number of factors of 5 in each term k^(2k).

For k = 1 to 25:
– k = 5: 5^10 contributes 10 factors of 5.
– k = 10: 10^20 = (2 × 5)^20 contributes 20 factors of 5.
– k = 15: 15^30 = (3 × 5)^30 contributes 30 factors of 5.
– k = 20: 20^40 = (2^2 × 5)^40 contributes 40 factors of 5.
– k = 25: 25^50 = (5^2)^50 contributes 100 factors of 5.

Summing these contributions:
10 + 20 + 30 + 40 + 100 = 200

Multiple of 2 is definitely more.
So, Answer is 200