JEE Main 2025 Jan 29 Shift 2, Mathematics Q17: Quadratic Equations

We need to find the remainder when $7^{103}$ is divided by $23$.

Note: $a\equiv b(\mathrm{mod}23)$ means that when $a$ and $b$ are divided by $23$, both leave the same remainder.

For example, $49\equiv 3(\mathrm{mod}23)$ because $49=23\times 2+3$.

Now, compute powers of $7$ modulo $23$:

$7^2=49\equiv 3(\mathrm{mod}23)$

$7^4=(7^2{)}^2\equiv 3^2=9(\mathrm{mod}23)$

$7^8=(7^4{)}^2\equiv 9^2=81\equiv 12(\mathrm{mod}23)$

Now,

$7^{11}=7^8\cdot 7^2\cdot 7\equiv 12\cdot 3\cdot 7=252\equiv -1(\mathrm{mod}23)$

Here, $252=23\times 11-1$, so the remainder is $-1$. Since remainders are usually taken as non-negative, add $23$:

$-1+23=22\Rightarrow -1\equiv 22(\mathrm{mod}23)$

Since $99=11\times 9$,

$7^{99}=(7^{11}{)}^9\equiv (-1{)}^9=-1(\mathrm{mod}23)$

Also, $103=99+4$.

$\therefore 7^{103}=7^{99}\cdot 7^4\equiv (-1)\cdot 9=-9\equiv 14(\mathrm{mod}23)$

Similarly, $-9+23=14$, so $-9$ and $14$ leave the same remainder when divided by $23$.

$\boxed{\text{Remainder}=14}$

Shortcut (Modular Arithmetic): Although modular arithmetic is not a separate JEE syllabus topic, its basic properties are frequently used as a faster method to solve remainder problems.