JEE 2023 — Mathematics PYQ

Solving:

1. $2023=35\times 57+28⟹2023\equiv 28(\mathrm{mod}35)\equiv -7(\mathrm{mod}35)$

2. Let $X=(2023{)}^{2023}(\mathrm{mod}35)$.

3. Using Chinese Remainder Theorem:

   • $(\mathrm{mod}7)$: $2023$ is a multiple of $7$ ($7\times 289$), so $2023\equiv 0(\mathrm{mod}7)$.

     Thus, $(2023{)}^{2023}\equiv 0(\mathrm{mod}7)$.

   • $(\mathrm{mod}5)$: $2023\equiv 3(\mathrm{mod}5)$.

     $3^1\equiv 3$, $3^2\equiv 4$, $3^3\equiv 2$, $3^4\equiv 1(\mathrm{mod}5)$.

     $2023=4\times 505+3⟹3^{2023}\equiv 3^3\equiv 27\equiv 2(\mathrm{mod}5)$.

4. Find $X$ such that $X\equiv 0(\mathrm{mod}7)$ and $X\equiv 2(\mathrm{mod}5)$:

   • Multiples of $7$: $7,14,21,28,\dots$

   • Check $(\mathrm{mod}5)$: $7\equiv 2(\mathrm{mod}5)$.

   • Therefore, $X=7$.

Answer: 7