The digit in the unit's place of $[(251)^{98} + (21)^{29} - (106)^{100} + (705)^{35} - 16^4 + 259]$ is

### Concept & Logic The numbers $1, 5,$ and $6$ have a cyclicity of $1$. This means that any number ending in $1, 5,$ or $6$ will always retain that same unit digit when raised to any positive integer power. ### Step-by-Step Solution **Given:** Evaluate the unit digit of $[(251)^{98} + (21)^{29} - (106)^{100} + (705)^{35} - 16^4 + 259]$. **Calculation:** * Extract the unit digit for each term based on its base: * $(251)^{98} \rightarrow$ base ends in $1$, so unit digit is $1$. * $(21)^{29} \rightarrow$ base ends in $1$, so unit digit is $1$. * $(106)^{100} \rightarrow$ base ends in $6$, so unit digit is $6$. * $(705)^{35} \rightarrow$ base ends in $5$, so unit digit is $5$. * $16^4 \rightarrow$ base ends in $6$, so unit digit is $6$. * $259 \rightarrow$ unit digit is $9$. * Substitute these unit digits back into the expression: $$1 + 1 - 6 + 5 - 6 + 9$$ * Add the positive values together and the negative values together: * Positive sum: $1 + 1 + 5 + 9 = 16 \rightarrow$ unit digit $6$. * Negative sum: $-6 - 6 = -12 \rightarrow$ unit digit $2$. * Subtract the negative unit sum from the positive unit sum: $$6 - 2 = 4$$ ### Exam Strategy & Shortcut Scan the expression for bases ending in $0, 1, 5,$ or $6$ because their unit digits never change. Since every single term in this expression falls into that category (or is a flat number like $259$), you can instantly write down $1 + 1 - 6 + 5 - 6 + 9$ and solve it mentally in under 15 seconds. ### Common Pitfall A major trap in long polynomial additions/subtractions is ending up with a negative unit digit (e.g., if you evaluated left-to-right as $2 - 6 = -4$). If you ever hit a negative intermediate unit digit, you must "borrow $10$" (so $-4$ becomes $10 - 4 = 6$). Grouping all positives and all negatives first avoids this confusion. ### Final Answer **Therefore, the correct answer is 4.**