Question

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

• A. $s<6$
• B. $6\le s<140$
• C. $140\le s<148$
• D. $s\ge 148$

Answer 1

The correct option is B $6\le s<140$

Let $N={15}^2\times 5^{18}=3^2\times 5^{20}=9\times 5^{20}$
${\log }_{10}(9\times 5^{20})=2{\log }_{10}3+20{\log }_{10}5=2\left(0.4771\right)+20{\log }_{10}^{\frac{10}{2}}=2\left(0.4771\right)+20(1-0.3010)=14.93$
The characterstic will be 14.
Hence the number will have 15 digits.
maximum when all digits are 9 will be $=9\times 15=135$
As we can see the last digit will be 5,
so minimum will be $\ge 6$

Hence the value of $s$ will lie in $6\le s<140$