$(7^{19} + 2)$ is divided by $6$. The remainder is

### Concept & Logic This problem uses the **Additive Property of Remainders**. If you are dividing a sum of two terms by a divisor, you can find the remainder of each term individually and then add those remainders together. $$ (A + B) \pmod D = [A \pmod D + B \pmod D] \pmod D $$ ### Step-by-Step Solution * **Given:** * Expression: $7^{19} + 2$ * Divisor: $6$ * **Calculation:** * Break the expression into two parts and divide each by $6$. * **Part 1:** Find the remainder of $7^{19} \div 6$. * Since $7$ is exactly $1$ more than the divisor $6$, we can write $7 = 6 + 1$. * $7^{19} \pmod 6 = (6 + 1)^{19} \pmod 6$. * As per the remainder theorem, this simply becomes $1^{19} = 1$. * **Part 2:** Find the remainder of $2 \div 6$. * Since $2$ is smaller than $6$, the remainder is simply $2$. * **Combine:** Add the individual remainders together. $$ \text{Total Remainder} = 1 + 2 = 3 $$ ### Exam Strategy & Shortcut Whenever the base is exactly $1$ more than the divisor (like $7$ and $6$), the remainder of that term to *any* power is always exactly $1$. You can mentally skip all the math: $1 + 2 = 3$. This should take no more than 3 seconds to solve on test day. ### Common Pitfall A careless mistake is trying to evaluate the power or add the $2$ to the base before dealing with the exponent (e.g., falsely thinking it acts like $9^{19}$). Always isolate terms separated by a plus or minus sign when applying the remainder theorem. ### Final Answer **Therefore, the correct answer is 3.**