If $(67^{67} + 67)$ is divided by $68$, the remainder is

### Concept & Logic This relies heavily on modular arithmetic, specifically evaluating each term in an additive sequence independently against the divisor. $$ (A + B) \pmod C \equiv (A \pmod C + B \pmod C) \pmod C $$ ### Step-by-Step Solution * **Given:** Find the remainder when $(67^{67} + 67)$ is divided by $68$. * **Split the expression:** Apply the modulo $68$ operation to both terms independently. Term 1: $67^{67} \pmod{68}$ Term 2: $67 \pmod{68}$ * **Evaluate Term 1 (The Exponent):** Using negative remainders, $67$ is $(68 - 1)$. $$67 \equiv -1 \pmod{68}$$ $$(-1)^{67} = -1$$ So, the remainder for the first term is $-1$. * **Evaluate Term 2 (The Constant):** The number $67$ divided by $68$ simply leaves a remainder of $67$. Alternatively, we can use the negative remainder $-1$ to make addition easier. * **Combine Remainders:** Add the remainders of the two terms together. $$\text{Total Remainder} = -1 + 67 = 66$$ ### Exam Strategy & Shortcut Use negative remainders for all terms close to the divisor. $67 \pmod{68}$ is $-1$. Substitute $-1$ into the entire expression: $(-1)^{67} + (-1) = -1 - 1 = -2$. Convert the negative remainder back to positive: $68 - 2 = 66$. This avoids large numbers entirely. ### Common Pitfall Students often correctly evaluate $67^{67} \pmod{68}$ as $-1$ but forget to carry down the second term ($+ 67$) from the original problem, leading them to incorrectly select $67$ as the final answer. ### Final Answer Therefore, the correct answer is **66**.