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JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Binomial Theorem

3 \times 7^{22} + 2 \times 1 0^{22} - 44

when divided by 18 leaves the remainder ______.

Answer: 15

Solution

Since

7^{3} = 343 \equiv 1 (mod 18)

,

7^{22} = (7^{3})^{7} \cdot 7 \equiv 7 (mod 18)

. Since

1 0^{2} = 100 \equiv 10 (mod 18)

,

1 0^{n} \equiv 10 (mod 18)

for all

n \geq 1

, so

1 0^{22} \equiv 10

. Then

3 \cdot 7^{22} + 2 \cdot 1 0^{22} - 44 \equiv 3 (7) + 2 (10) - 44 = - 3 \equiv 15 (mod 18) .

The remainder is

15

.

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Integer answer

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