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

Option: 1

23


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

x^{\frac{1}{50}}=12 \quad y^{\frac{1}{50}}=18

Remainder when \mathrm{x}+\mathrm{y}  is division by 25 .

\mathrm{x=12^{50} \quad y=18^{50}}

\mathrm{\mathrm{x}+\mathrm{y}=12^{50}+18^{50}}
\mathrm{=6^{50}\left(2^{50}+3^{50}\right)}
\mathrm{=(5+1)^{50}\left(\left(2^{2}\right)^{25}+\left(3^{2}\right)^{25}\right)}
\mathrm{=\left(25 \lambda_{1}+1\right)\left((5-1)^{25}+(10-1)^{25}\right)}
\mathrm{=\left(25 \lambda_{1}+1\right)\left(25\left(\lambda_{2}+\lambda_{3}\right)-2\right)}
\mathrm{=\left(25 \lambda_{1}+1\right)(25 \mathrm{~K}-2)}
\mathrm{\Rightarrow 25 \lambda_{1} .25 \mathrm{~K}-50 \lambda_{1}+25 \mathrm{~K}-2}
\mathrm{\Rightarrow 25 \mathrm{n}_{1}-2}
\mathrm{\Rightarrow 25 \mathrm{n}_{2}+23}

Remainder \mathrm{=23}

Posted by

seema garhwal

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