JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Binomial Theorem

Group the terms as $\left(25^{190}-8^{190}\right)-\left(19^{190}-2^{190}\right)$. Since $a^n-b^n$ is divisible by $a-b$ and $25-8=19-2=17$, each bracket is divisible by $17$, so the whole expression is divisible by $17$. The parity is odd$-$odd$-$even$+$even, which is even, so the expression is also divisible by $2$, and hence by $2\cdot 17=34$.

For $14=2\cdot 7$, work modulo $7$: $25\equiv 4,19\equiv 5,8\equiv 1,2\equiv 2$. With $190\equiv 1(\mathrm{mod}3)$ and $190\equiv 4(\mathrm{mod}6)$, $4^{190}\equiv 4,5^{190}\equiv 2,1^{190}\equiv 1,2^{190}\equiv 2$, so the expression $\equiv 4-2-1+2=3(\mathrm{mod}7)$. It is not divisible by $7$, hence not by $14$.

Therefore the expression is divisible by $34$ but not by $14$.