Question

If $a,b,c$ are positive numbers such that $a^{{\log }_{3}7}=27,b^{{\log }_{7}11}=49,c^{{\log }_{11}25}=\sqrt{11}$, then the sum of digits of $S=a^{({\log }_{3}7{)}^2}+b^{({\log }_{7}11{)}^2}+c^{({\log }_{11}25{)}^2}$ is

• A. 15
• B. 17
• C. 19
• D. 21

Answer 1

The correct option is C 19

$a^{{\log }_{3}7}=27,b^{{\log }_{7}11}=49,c^{{\log }_{11}25}=\sqrt{11}$
$S=(a^{{\log }_{3}7}{)}^{{\log }_{3}7}+{\left(b^{{\log }_{7}11}\right)}^{{\log }_{7}11}+{\left(c^{{\log }_{11}25}\right)}^{{\log }_{11}25}$
$={\left(27\right)}^{{\log }_{3}7}+{\left(49\right)}^{{\log }_{7}11}+{\left(\sqrt{11}\right)}^{{\log }_{11}25}$
$=3^{3{\log }_{3}7}+7^{2{\log }_{7}11}+{11}^{1/2{\log }_{11}25}$
$=7^3+{11}^2+{25}^{1/2}$
$=469$
Sum of digits $=19$
Hence, C is correct.