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

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Basic of Logarithms

normal

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

$S < 6$

$6 \leq S < 140$

$140 \leq S < 148$

$S \geq 148$

(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

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