JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let the number $(22)^{2022} + (2022)^{22}$ leave the remainder $\alpha$ when divided by $3$ and $\beta$ when divided by $7$ . Then $(\alpha^2 + \beta^2)$ is equal to:

Choose the correct answer:

Explanation

Solution:

1. $\alpha$ nikalna (Remainder when divided by 3):

(22)^{2022} + (2022)^{22} \pmod{3}

$22 \equiv 1 \pmod{3}$ , toh $(22)^{2022} \equiv 1^{2022} \equiv 1 \pmod{3}$ .
$2022$ is divisible by $3$ (digits ka sum $2+0+2+2=6$ hai), toh $(2022)^{22} \equiv 0 \pmod{3}$ .
$\alpha = 1 + 0 = 1$ .

2. $\beta$ nikalna (Remainder when divided by 7):

(22)^{2022} + (2022)^{22} \pmod{7}

$22 \equiv 1 \pmod{7}$ , toh $(22)^{2022} \equiv 1^{2022} \equiv 1 \pmod{7}$ .
$2022 \pmod{7}$ : $2022 = 7 \times 288 + 6$ . So, $2022 \equiv 6 \equiv -1 \pmod{7}$ .
$(-1)^{22} \equiv 1 \pmod{7}$ .
$\beta = 1 + 1 = 2$ .

3. Final Calculation:

\alpha^2 + \beta^2 = 1^2 + 2^2 = 1 + 4 = 5

Choose the correct answer:

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