Binomial Theorem – Remainder – JEE Main 29 Jan 2025 Shift 2

Solution:

We write the expression in terms of powers that are close to multiples of 23.
$$ 7^{103} = 7^1 \cdot 7^{102} = 7 \cdot (7^3)^{34} $$

We know that $7^3 = 343$. Dividing 343 by 23:
$$ 343 = 23 \times 15 – 2 $$
So, we can write $7^3 = (15 \cdot 23 – 2)$.

Substitute this into the expression:
$$ 7^{103} = 7 \cdot (15 \cdot 23 – 2)^{34} $$

Now, expand $(15 \cdot 23 – 2)^{34}$ using the Binomial Theorem $(x-y)^n$.
$$ (15 \cdot 23 – 2)^{34} = ^{34}C_0(15 \cdot 23)^{34} – ^{34}C_1(15 \cdot 23)^{33}(2) + \dots + ^{34}C_{34}(-2)^{34} $$
All terms in this expansion contain a factor of 23, except the last term.
$$ 7^{103} = 7 \cdot (23k + (-2)^{34}) \quad \text{where } k \text{ is an integer} $$

Taking modulo 23:
$$ 7^{103} \equiv 7 \cdot 2^{34} \pmod{23} $$

Now we apply the Binomial Theorem again to $2^{34}$. We look for a power of 2 close to a multiple of 23.
Consider $2^{11} = 2048$.
Dividing 2048 by 23:
$$ 2048 = 23 \times 89 + 1 $$

We rewrite $2^{34}$:
$$ 2^{34} = 2^1 \cdot 2^{33} = 2 \cdot (2^{11})^3 $$
Substitute $2^{11} = (89 \cdot 23 + 1)$:
$$ 2^{34} = 2 \cdot (89 \cdot 23 + 1)^3 $$

Expand $(89 \cdot 23 + 1)^3$ using Binomial Theorem:
$$ (89 \cdot 23 + 1)^3 = ^{3}C_0(89 \cdot 23)^3 + \dots + ^{3}C_3(1)^3 $$
Modulo 23, all terms vanish except the last one:
$$ 2^{34} \equiv 2 \cdot (1)^3 \equiv 2 \pmod{23} $$

Finally, substitute this back into the expression for $7^{103}$:
$$ 7^{103} \equiv 7 \cdot 2 \pmod{23} $$
$$ 7^{103} \equiv 14 \pmod{23} $$

Ans. (1)