Unit's digit in $(784)^{126} + (784)^{127}$ is

### Concept & Logic For numbers ending in $4$ or $9$, the cyclicity is simply $2$. You don't need to divide by $4$; you only need to look at whether the exponent is odd or even to determine the unit digit. ### Step-by-Step Solution **Given:** Evaluate the unit digit of $(784)^{126} + (784)^{127}$. **Calculation:** * The base for both terms has a unit digit of $4$. We only care about powers of $4$. * The cyclicity rule for $4$ is: * $4^{\text{odd power}}$ ends in $4$ (like $4^1 = 4$). * $4^{\text{even power}}$ ends in $6$ (like $4^2 = 16$). * **Term 1:** $(784)^{126}$ has an even exponent ($126$). Thus, its unit digit is $6$. * **Term 2:** $(784)^{127}$ has an odd exponent ($127$). Thus, its unit digit is $4$. * **Addition:** Add the two derived unit digits: $$6 + 4 = 10$$ * The final unit digit of the sum $10$ is $0$. ### Exam Strategy & Shortcut Notice that $a^n + a^{n+1}$ can be factored. $(784)^{126} \times (1 + 784) = (784)^{126} \times 785$. Any even number (which $(784)^{126}$ definitely is) multiplied by a number ending in $5$ (like $785$) will always produce a trailing zero. This logic completely bypasses cyclicity calculations! ### Common Pitfall Applying the "divide by $4$" cyclicity rule to bases ending in $4$ or $9$ isn't *wrong*, but it is an inefficient use of time. Recognize the odd/even shortcut to save precious seconds on the exam. ### Final Answer **Therefore, the correct answer is 0.**