UPSC CDS 2 2019 Math Paper

Concept:The largest $n$ such that $10^n$ divides the given expression is the minimum exponent of the common factors $2$ and $5$ in the prime factorization of the product.Explanation:First, factor each term completely:$6^{23} = (2 \times 3)^{23} = 2^{23} \times 3^{23}$$75^{9} = (3 \times 5^2)^{9} = 3^9 \times 5^{18}$$105^{2} = (3 \times 5 \times 7)^{2} = 3^2 \times 5^2 \times 7^2$Now multiply all the prime factors:$2^{23} \times 3^{23+9+2} \times 5^{18+2} \times 7^2$$= 2^{23} \times 3^{34} \times 5^{20} \times 7^2$Since $10^n = (2 \times 5)^n = 2^n \times 5^n$, the exponent of $n$ is limited by the smaller exponent among $2$ and $5$ in the product.Here, exponent of $2$ is $23$, exponent of $5$ is $20$. The minimum is $20$.Thus, the largest $n$ such that $10^n$ divides the expression is $20$.Answer:$n = 20$ (Option A)