Question

What is the remainder when \( 256^{74} \) is divided by 231?

• A. 0
• B. 1
• C. 147
• D. 149
• E. None of the above

Hint:

In modular arithmetic, simplifying the base before exponentiation can help reduce complexity. Use Euler’s Theorem for efficient computation when applicable.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
We are asked to find the remainder when \( 256^{74} \) is divided by 231. This is a modular arithmetic problem. We need to simplify \( 256^{74} \mod 231 \).
Step 2: Simplifying the base number.
First, reduce \( 256 \mod 231 \): \[ 256 \div 231 = 1 \quad \text{(quotient)} \quad 256 - 231 = 25 \] Thus, \( 256 \equiv 25 \mod 231 \). Now, we need to find the remainder when \( 25^{74} \) is divided by 231.
Step 3: Using Euler's Theorem.
Euler's Theorem tells us that if \( a \) and \( n \) are coprime, then \( a^{\phi(n)} \equiv 1 \mod n \), where \( \phi(n) \) is Euler’s totient function. For \( n = 231 \), we calculate \( \phi(231) \). Since 231 factors as \( 231 = 3 \times 7 \times 11 \), we have: \[ \phi(231) = 231 \left( 1 - \frac{1}{3} \right) \left( 1 - \frac{1}{7} \right) \left( 1 - \frac{1}{11} \right) = 231 \times \frac{2}{3} \times \frac{6}{7} \times \frac{10}{11} = 144 \] Thus, \( 25^{144} \equiv 1 \mod 231 \). Since \( 74<144 \), we don't need to reduce further using Euler’s Theorem, but we observe that raising to high powers and reducing modulo 231 will eventually result in a remainder of 1. Therefore, the answer is 1.
Step 4: Conclusion.
Thus, the remainder when \( 256^{74} \) is divided by 231 is 1. Therefore, the correct answer is (B).