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JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Binomial Theorem

The remainder, when

7^{103}

is divided by 23 , is equal to:

(A)14✓ Correct
(B)9
(C)17
(D)6

Answer: (A)

Solution

7^{103} = 7 \cdot 7^{102} = 7 \cdot 34 3^{34}

Since

343 = 15 \times 23 - 2 \equiv - 2 (mod 23)

,

7^{103} \equiv 7 \cdot (- 2)^{34} = 7 \cdot 2^{34} (mod 23)

Now

2^{34} = 4 \cdot (2^{8})^{4} = 4 \cdot 25 6^{4}

, and

256 = 11 \times 23 + 3 \equiv 3 (mod 23)

, so

25 6^{4} \equiv 3^{4} = 81 \equiv 81 - 69 = 12 (mod 23)

7 \cdot 2^{34} \equiv 7 \cdot 4 \cdot 12 = 336 \equiv 336 - 14 \times 23 = 336 - 322 = 14 (mod 23)

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option

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