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

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}$12×24×36×48×...×2550. We need to identify terms that contribute factors of 5:

• From $5^{10}$510: This term contributes $5^{10}$510 to the product.
• From $10^{20}$1020: This term is $(2 \times 5)^{20} = 2^{20} \times 5^{20}$(2×5)20=220×520, contributing $5^{20}$520.
• From $15^{30}$1530: This term is $(3 \times 5)^{30} = 3^{30} \times 5^{30}$(3×5)30=330×530, contributing $5^{30}$530.
• From $20^{40}$2040: This term is $(4 \times 5)^{40} = (2^2 \times 5)^{40} = 2^{80} \times 5^{40}$(4×5)40=(22×5)40=280×540, contributing $5^{40}$540.
• From $25^{50}$2550: This term is $(5^2)^{50} = 5^{100}$(52)50=5100, contributing $5^{100}$5100.

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

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

Thus, the product contains $5^{200}$5200 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.