Let S be the sum of the digits of the number | vedclass

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(B) Given number $n = 15^2 	imes 5^{18}$.$n = (3 	imes 5)^2 	imes 5^{18} = 3^2 	imes 5^2 	imes 5^{18} = 9 	imes 5^{20}$.To find the number of di

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$6 \leq S < 140$

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(B) Given number $n = 15^2 	imes 5^{18}$.$n = (3 	imes 5)^2 	imes 5^{18} = 3^2 	imes 5^2 	imes 5^{18} = 9 	imes 5^{20}$.To find the number of digits,we calculate $\log_{10} n = \log_{10} (9 	imes 5^{20}) = \log_{10} 9 + 20 \log_{10} 5$.Using $\log_{10} 3 \approx 0.4771$ and $\log_{10} 5 \approx 0.6990$:$\log_{10} n = 2 	imes 0.4771 + 20 	imes 0.6990 = 0.9542 + 13.98 = 14.9342$.Since the characteristic is $14$,the number $n$ has $14 + 1 = 15$ digits.The maximum possible sum of digits for a $15$-digit number is $9 	imes 15 = 135$. However,$n = 9 	imes 5^{20}$ ends in $5$ (since $5^k$ ends in $5$ for $k \geq 1$),so the last digit is $5$.The sum of digits $S$ satisfies $S \leq 9 	imes 14 + 5 = 126 + 5 = 131$.Since the number is not zero,$S \geq 1$ (actually $S \geq 6$ as the sum of digits of a multiple of $9$ is a multiple of $9$,and $131$ is not a multiple of $9$,but $S$ must be at least $6$ based on the options provided).Thus,$6 \leq S < 140$.

Let $S$ be the sum of the digits of the number $15^2 \times 5^{18}$ in base $10$. Then,

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