JEE 2023 — Mathematics PYQ

Solution:

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

$(x+y)(\mathrm{mod}25)=(12^{50}+18^{50})(\mathrm{mod}25)$

$12^2=144\equiv -6(\mathrm{mod}25)$

$12^{10}=(12^2{)}^5\equiv (-6{)}^5=-7776\equiv -1(\mathrm{mod}25)$

$12^{50}=(12^{10}{)}^5\equiv (-1{)}^5=-1(\mathrm{mod}25)$

$18\equiv -7(\mathrm{mod}25)⟹18^{50}\equiv (-7{)}^{50}=7^{50}(\mathrm{mod}25)$

$7^2=49\equiv -1(\mathrm{mod}25)$

$7^{50}=(7^2{)}^{25}\equiv (-1{)}^{25}=-1(\mathrm{mod}25)$

Remainder $=-1+(-1)=-2$

Final Remainder $=-2+25=23$