QJEEM23Ma554e: JEE Main 2023 Mathematics - Class 11 JEEM23Ma554e Solution: root of a number is 12 and root of another number is 18 . Then…

Given ${x^{{1 \over {50}}}} = 12 \Rightarrow x = {12^{50}}$

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

$12\equiv13$ (Mod 25)

$12^2\equiv19$ (Mod 25)

$12^3\equiv-3$ (Mod 25)

$12^9\equiv-2$ (Mod 25)

$12^{10}\equiv-1$ (Mod 25)

$12^{50}\equiv-1$ (Mod 25) ..... (i)

Now

$18\equiv7$ (Mod 25)

$18^2\equiv-1$ (Mod 25)

$18^{-50}\equiv-1$ (Mod 25) ..... (ii)

$\therefore$ $12^{50}+18^{50}\equiv-2$ (Mod 25)

$\equiv23$ (Mod 25)

$\therefore$ Answer = 23