$(9^6 + 1)$ when divided by $8$, would leave a remainder of

### Concept & Formula This question utilizes the Remainder Theorem and Modular Arithmetic. When a number $(x + a)^n$ is divided by $x$, the remainder is strictly dependent on $a^n$. $$(x + a)^n \pmod x \equiv a^n \pmod x$$ ### Step-by-Step Solution * **Given:** We need to find the remainder of $(9^6 + 1)$ divided by 8. * **Rewrite the base:** Express the base $9$ in terms of the divisor $8$. $$9 = 8 + 1$$ So, the expression becomes $((8 + 1)^6 + 1) \div 8$. * **Apply the Remainder Theorem:** The remainder of $(8 + 1)^6$ when divided by 8 is simply the remainder of $1^6$. $$1^6 = 1$$ * **Calculate total remainder:** We must not forget the "+ 1" present in the original expression. $$\text{Total Remainder} = 1 + 1 = 2$$ ### Exam Strategy & Shortcut Use modular arithmetic directly. Since $9 \equiv 1 \pmod 8$, any power of $9$ will also be congruent to $1 \pmod 8$. So, $(9^6 + 1) \pmod 8 \equiv (1^6 + 1) \pmod 8 \equiv 2$. This can be solved in under 5 seconds without writing anything down. ### Common Pitfall The most common error is finding the remainder of $9^6$ (which is 1) and forgetting to add the trailing $+1$ from the original question, leading a student to incorrectly choose option (b). ### Final Answer Therefore, the correct answer is **2**.