50 th  root of a number x is 12 and 50 th  root of another number y is 18   . Then the…

Given:

$x^{\frac{1}{50}}=12\Rightarrow x={12}^{50}$

$y^{\frac{1}{50}}=18\Rightarrow y={18}^{50}$

So,

$x+y={12}^{50}+{18}^{50}$

$\Rightarrow \frac{x+y}{25}=\frac{{12}^{50}}{25}+\frac{{18}^{50}}{25}$

$\Rightarrow \frac{x+y}{25}=\frac{{\left({12}^3\right)}^{16}{12}^2}{25}+\frac{{\left({18}^2\right)}^{25}}{25}$

$\Rightarrow \frac{x+y}{25}=\frac{{\left(1728\right)}^{16}{12}^2}{25}+\frac{{\left(325-1\right)}^{25}}{25}$

$\Rightarrow \frac{x+y}{25}=\frac{{\left(1725+3\right)}^{16}{12}^2}{25}+\frac{{\left(325-1\right)}^{25}}{25}$

So, remainder is

$\frac{3^{16}{12}^2}{25}-1=\frac{{\left(3^3\right)}^5\times 3\times 144}{25}-1$

$=\frac{{\left(25+2\right)}^5\times 3\times 144}{25}-1$

$=\frac{{\left(25+2\right)}^5\times 3\times \left(150-6\right)}{25}-1$

$=\frac{2^5\times 3\times \left(-6\right)}{25}-1$

$=\frac{\left(25+7\right)\times 3\times \left(-6\right)}{25}-1$

$=\frac{\left(21\right)\times \left(-6\right)}{25}-1$

$=\frac{\left(25-4\right)\times \left(-6\right)}{25}-1$

$=\frac{\left(-4\right)\times \left(-6\right)}{25}-1$

$=24-1=23$

Hence, the remainder is $23$