IPMAT Indore 2025 (MCQ) - The remainder when 11^1011 + 1011^11 is divided by 9 is | PYQs + Solutions

IPMAT Indore 2025

Number System

Remainder

Medium

The remainder when $11^{1011} + 1011^{11}$ is divided by $9$ is

✅ Correct Option: 2

We know that:

11^{1011} \equiv 2^{1011} \pmod{9}

Let's examine the pattern of powers of

2

modulo

9

\newline

2^1 \equiv 2 \pmod{9}

\newline

2^2 \equiv 4 \pmod{9}

\newline

2^3 \equiv 8 \pmod{9}

\newline

2^4 \equiv 16 \equiv 7 \pmod{9}

\newline

2^5 \equiv 14 \equiv 5 \pmod{9}

\newline

2^6 \equiv 10 \equiv 1 \pmod{9}

We see the pattern repeats every

6

powers, so we need to find

1011 \mod 6

1011 = 6 \times 168 + 3

Therefore,

2^{1011} \equiv 2^3 \equiv 8 \pmod{9}

For

1011^{11}

, we first determine

1011 \mod 9

Using the digit sum rule:

1 + 0 + 1 + 1 = 3

1011 \equiv 3 \pmod{9}

, which means

1011^{11} \equiv 3^{11} \pmod{9}

Examining powers of

3

modulo

9

\newline

3^1 \equiv 3 \pmod{9}

\newline

3^2 \equiv 9 \equiv 0 \pmod{9}

Since

3^2 \equiv 0 \pmod{9}

, any higher power will also be

0

modulo

9

Therefore,

3^{11} \equiv 0 \pmod{9}

Finally,

11^{1011} + 1011^{11} \equiv 8 + 0 \equiv 8 \pmod{9}

The remainder is

\boxed{8}

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