JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Binomial Theorem

Here $x=12^{50}$ and $y=18^{50}$. Writing $12=15-3$ and $18=15+3$, the expansions of $(15-3{)}^{50}$ and $(15+3{)}^{50}$ add to $2{\sum }_{k\text{ even}}\left(\genfrac{}{}{0ex}{}{50}{k}\right)15^{50-k}{3}^k$. Modulo $25$ every term that still carries a factor $15^2=225$ vanishes, and since $50-k$ is even it is either $0$ or at least $2$, so only the $k=50$ term, $3^{50}$, survives. Now $3^{10}=59049=25\times 2361+24$, so $3^{10}\equiv -1(\mathrm{mod}25)$, and hence $3^{50}=(3^{10}{)}^5\equiv -1$. Therefore $x+y\equiv 2(-1)=-2\equiv 23(\mathrm{mod}25)$, so the remainder obtained on dividing $x+y$ by $25$ is $23$.