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

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.