JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The remainder, when $7^{103}$ is divided by $17$ , is ________.

Choose the correct answer:

Explanation

Step 1: Fermat's Little Theorem ka upyog

Fermat's Little Theorem ke anusar, yadi $p$ ek prime number hai aur $a$ , $p$ se divide nahi hota, toh:

a^{p-1} \equiv 1 \pmod{p}

Yahan $a = 7$ aur $p = 17$ hai (jo ki ek prime number hai).

Isliye,

7^{17-1} \equiv 1 \pmod{17}

\Rightarrow 7^{16} \equiv 1 \pmod{17}

Step 2: Power ko simplify karna

Humein $7^{103}$ chahiye. Hum $103$ ko $16$ ke multiples mein likhenge:

103 = 16 \times 6 + 7

Toh,

7^{103} = (7^{16})^6 \times 7^7

Step 3: Modular Arithmetic lagana

Kyunki $7^{16} \equiv 1 \pmod{17}$ , isliye:

7^{103} \equiv (1)^6 \times 7^7 \pmod{17}

7^{103} \equiv 7^7 \pmod{17}

Step 4: $7^7 \pmod{17}$ ki value nikalna

$7^1 \equiv 7 \pmod{17}$
$7^2 = 49 = (17 \times 2) + 15 \equiv 15 \equiv -2 \pmod{17}$
$7^4 \equiv (-2)^2 = 4 \pmod{17}$
$7^6 = 7^2 \times 7^4 \equiv (-2) \times 4 = -8 \pmod{17}$
$7^7 = 7^6 \times 7^1 \equiv (-8) \times 7 = -56 \pmod{17}$

Ab $-56$ ko $17$ se divide karte hain:

$-56 = 17 \times (-4) + 12$

Isliye,

-56 \equiv 12 \pmod{17}

Final Answer:

Remainder jab $7^{103}$ ko $17$ se divide kiya jata hai, toh 12 aata hai.