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Binomial Theorem question

2022 · 25 Jul · Shift 2 · Q26

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Binomial Theorem
2022 · 25 Jul · Shift 2 · Q26

Binomial Theorem question

2022 · 25 Jul · Shift 2 · Q26

JEE MainMathematicsBinomial TheoremMCQ+4 / −1

The remainder when

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

is divided by 9 is

A
1
B
4
C
6
D
8

View written solutionFree

Correct answer: D

We need the remainder of $11^{1011}+1011^{11}$ when divided by $9$ .
Reduce each base modulo $9$ : $11 \equiv 2 \pmod{9}$ $1011 \equiv 3 \pmod{9}$ because $1011=9\cdot 112+3$ .

So, $11^{1011}+1011^{11} \equiv 2^{1011}+3^{11} \pmod{9}.$

Now evaluate $2^{1011} \pmod{9}$ .

The powers of $2$ modulo $9$ cycle as:

\quad 2^2\equiv 4, \quad 2^3\equiv 8, \quad 2^4\equiv 16\equiv 7, \quad 2^5\equiv 14\equiv 5, \quad 2^6\equiv 10\equiv 1 \pmod{9}.$$ So the cycle length is $6$. Now, $$1011 \equiv 3 \pmod{6}$$ therefore, $$2^{1011} \equiv 2^3 \equiv 8 \pmod{9}.$$ 4. Evaluate $3^{11} \pmod{9}$. Since $$3^2=9 \equiv 0 \pmod{9},$$ any higher power of $3$ is also divisible by $9$. Hence, $$3^{11} \equiv 0 \pmod{9}.$$ 5. Therefore, $$2^{1011}+3^{11} \equiv 8+0 \equiv 8 \pmod{9}.$$ So the remainder is $$\boxed{8}.$$ 6. Comparing with the stored correct answer: - Derived answer: $8$ - Stored correct answer: D ($8$) They match.

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