50^ th root of a number x is 12 and 50^ th root of another number y is 18 . Then the remainder obtained on dividing (x+y) by 25 is .

Given:

$${x}^{\frac{1}{50}}=12\Rightarrow x={12}^{50}$$x501​=12⇒x=1250

$${y}^{\frac{1}{50}}=18\Rightarrow y={18}^{50}$$y501​=18⇒y=1850

So,

$$x+y={12}^{50}+{18}^{50}$$x+y=1250+1850

$$\Rightarrow \frac{x+y}{25}=\frac{{12}^{50}}{25}+\frac{{18}^{50}}{25}$$⇒25x+y​=251250​+251850​

$$\Rightarrow \frac{x+y}{25}=\frac{{({12}^{3})}^{16}{12}^{2}}{25}+\frac{{({18}^{2})}^{25}}{25}$$⇒25x+y​=25(123)16122​+25(182)25​

$$\Rightarrow \frac{x+y}{25}=\frac{{(1728)}^{16}{12}^{2}}{25}+\frac{{(325-1)}^{25}}{25}$$⇒25x+y​=25(1728)16122​+25(325−1)25​

$$\Rightarrow \frac{x+y}{25}=\frac{{(1725+3)}^{16}{12}^{2}}{25}+\frac{{(325-1)}^{25}}{25}$$⇒25x+y​=25(1725+3)16122​+25(325−1)25​

So, remainder is

$$\frac{{3}^{16}{12}^{2}}{25}-1=\frac{{({3}^{3})}^{5}\times 3\times 144}{25}-1$$25316122​−1=25(33)5×3×144​−1

$$=\frac{{(25+2)}^{5}\times 3\times 144}{25}-1$$=25(25+2)5×3×144​−1

$$=\frac{{(25+2)}^{5}\times 3\times (150-6)}{25}-1$$=25(25+2)5×3×(150−6)​−1

$$=\frac{{2}^{5}\times 3\times (-6)}{25}-1$$=2525×3×(−6)​−1

$$=\frac{(25+7)\times 3\times (-6)}{25}-1$$=25(25+7)×3×(−6)​−1

$$=\frac{(21)\times (-6)}{25}-1$$=25(21)×(−6)​−1

$$=\frac{(25-4)\times (-6)}{25}-1$$=25(25−4)×(−6)​−1

$$=\frac{(-4)\times (-6)}{25}-1$$=25(−4)×(−6)​−1

$$=24-1=23$$=24−1=23

Hence, the remainder is $23$23