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Basic of Logarithms
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Let $S$ be the sum of the digits of the number $15^2 \times 5^{18}$ in base $10$. Then,
A
$S < 6$
B
$6 \leq S < 140$
C
$140 \leq S < 148$
D
$S \geq 148$
(KVPY-2018)
Solution
(b)
Given number,
$n=15^2 \times 5^{18}$
$n=3^2 \times 5^2 \times 5^{18}$
$n=9 \times 5^{20}$
Taking log base 10 both side
$\log _{10} n=\log _{10} 9+\log _{10} 5^{20}$
$=2 \log _{10} 3+20 \log _{10} 5$
$=2 \times 0.4771+20 \times(1-0.3010)$
$=14$ characters value
Hence, the number have $15$ digits $S=$ Sum of digits of the number Now, $n$ has last digit is $5 .$
$\therefore$ Minimum value of $S=1+5=6$
Maximum value of $S=9 \times 14+5$
$=126+5=131$
$\therefore 6 \leq S < 140$
Standard 11
Mathematics