50 ^ th root of a number x is 12 and 50 ^ th root of another number...

Given:

$x\dfrac{1}{50} = 12 ⇒ x = {12}^{50}$

$y\dfrac{1}{50} = 18 ⇒ y = {18}^{50}$

So,

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

$⇒ \dfrac{x + y}{25} = \dfrac{{12}^{50}}{25} + \dfrac{{18}^{50}}{25}$

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

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

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

So, remainder is

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

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

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

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

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

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

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

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

$= 24 - 1 = 23$

Hence, the remainder is $23$