Question

The remainder when \( 11^{1011} + 1011^{11} \) is divided by 9 is

• A. 0
• B. 7
• C. 9
• D. 8

Hint:

When finding remainders of large powers \(a^b \pmod{m}\), first reduce the base \(a \pmod{m}\), then use the cyclicity of the powers of the new base to reduce the exponent \(b\). This simplifies the calculation significantly.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem requires finding the remainder of a large sum, which can be efficiently solved using modular arithmetic. We need to find the value of the expression modulo 9.
Step 2: Key Formula or Approach:
We will use the properties of congruences: If \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\), then \(a+c \equiv b+d \pmod{m}\) and \(a^k \equiv b^k \pmod{m}\). The divisibility rule for 9 states that a number is congruent to the sum of its digits modulo 9.
Step 3: Detailed Explanation:
We need to compute \( (11^{1011} + 1011^{11}) \pmod{9} \). We can compute each term separately. Term 1: \( 11^{1011} \pmod{9} \) First, find the remainder of the base: \( 11 \div 9 \) gives a remainder of 2. So, \( 11 \equiv 2 \pmod{9} \). Therefore, \( 11^{1011} \equiv 2^{1011} \pmod{9} \). Now we look for the cycle of powers of 2 modulo 9: \( 2^1 \equiv 2 \) \( 2^2 \equiv 4 \) \( 2^3 \equiv 8 \) \( 2^4 \equiv 16 \equiv 7 \) \( 2^5 \equiv 14 \equiv 5 \) \( 2^6 \equiv 10 \equiv 1 \) The cycle length is 6. We need to find the exponent 1011 modulo 6. \( 1011 \div 6 \): \( 1011 = 6 \times 168 + 3 \). So, \( 1011 \equiv 3 \pmod{6} \).
This means \( 2^{1011} \equiv 2^3 \pmod{9} \).
\( 2^3 = 8 \). So, \( 11^{1011} \equiv 8 \pmod{9} \).
Term 2: \( 1011^{11} \pmod{9} \) First, find the remainder of the base using the sum of digits rule: Sum of digits of 1011 is \( 1+0+1+1 = 3 \). So, \( 1011 \equiv 3 \pmod{9} \). Therefore, \( 1011^{11} \equiv 3^{11} \pmod{9} \). Now we look at powers of 3 modulo 9:
\( 3^1 \equiv 3 \) \( 3^2 = 9 \equiv 0 \) For any exponent \(k \ge 2\), \( 3^k = 3^{k-2} \cdot 3^2 \equiv 3^{k-2} \cdot 0 \equiv 0 \pmod{9} \). Since the exponent is 11, which is greater than 2, we have \( 1011^{11} \equiv 0 \pmod{9} \).
Final Sum: Now add the results for the two terms: \( 11^{1011} + 1011^{11} \equiv 8 + 0 \pmod{9} \) \( \equiv 8 \pmod{9} \) Step 4: Final Answer:
The remainder when \( 11^{1011} + 1011^{11} \) is divided by 9 is 8.