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JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Binomial Theorem

The remainder when

(11)^{1011} + (1011)^{11}

is divided by 9 is

(A)1
(B)4
(C)6
(D)8✓ Correct

Answer: (D)

Solution

Using modular arithmetic:

11 \equiv 2 (mod 9)

, so

1 1^{1011} \equiv 2^{1011} (mod 9)

. Since

2^{6} = 64 \equiv 1 (mod 9)

, we have

2^{1011} = 2^{6 \times 168 + 3} \equiv 2^{3} = 8 (mod 9)

. For the second term:

1011 = 112 \times 9 + 3 \equiv 3 (mod 9)

, so

101 1^{11} \equiv 3^{11} (mod 9)

. Since

3^{2} = 9 \equiv 0 (mod 9)

, we have

3^{11} = 3 \times (3^{2})^{5} \equiv 0 (mod 9)

. Therefore,

1 1^{1011} + 101 1^{11} \equiv 8 + 0 = 8 (mod 9)

.

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option

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