What is the remainder when $4^{96}$ is divided by 6?

### Concept & Formula This problem deals with the cyclicity and patterns of remainders (modular arithmetic). Specifically, we explore the properties of powers of 4 when divided by 6. $$4^n \equiv \text{constant remainder} \pmod 6$$ ### Step-by-Step Solution * **Step 1:** Let's analyze the remainder generated by small powers of 4 when divided by 6: * $4^1 = 4 \rightarrow 4 \div 6$ leaves a remainder of **4** * $4^2 = 16 \rightarrow 16 \div 6 = 2$ with a remainder of **4** * $4^3 = 64 \rightarrow 64 \div 6 = 10$ with a remainder of **4** * **Step 2:** Observe the clear rule: **Any positive integer power of 4 divided by 6 always yields a remainder of 4.** * **Step 3:** Apply this rule to $4^{96}$. Since $96$ is a positive integer, the remainder remains unchanged. ### Exam Strategy & Shortcut **Property Pattern:** Remember this standard number theory property: $$4^n \pmod 6 = 4 \quad \text{for all } n \in \mathbb{Z}^+$$ Since the exponent $96$ is a positive integer, you can instantly mark the answer as **4** without doing any complex modular reduction. ### Common Pitfall * **Over-complicating with Totient/Fermat Theorem:** Students sometimes try to apply Euler's totient function or Fermat's Little Theorem directly. However, because $4$ and $6$ are not co-prime ($\gcd(4,6) = 2$), standard basic theorems do not apply directly, leading to confusing breakdowns if the base relationship isn't noticed. ### Final Answer **Therefore, the correct answer is 4.**