Now, compute $ 3025 \mod 9 $. Since $ 55 \equiv 1 \mod 9 $, $ 55^2 \equiv 1^2 = 1 \mod 9 $.
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- Solution: The slope forms a right triangle with opposite side 1 and adjacent side $ \sqrt{3} $. Thus, $ \tan \theta = \frac{1}{\sqrt{3}} = \boxed{\frac{\sqrt{3}}{3}} $.Question: What is the remainder when the sum $ 1^3 + 2^3 + 3^3 + \dots + 10^3 $ is divided by 9?
- Solution: The formula for the sum of the first $ n $ cubes is $ \left( \frac{n(n+1)}{2} \right)^2 $. For $ n = 10 $, we compute:
- \left( \frac{10 \cdot 11}{2} \right)^2 = (55)^2 = 3025
- Thus, the remainder is $ \boxed{1} $.
- Question: What is the base-ten equivalent of the base-seven number $ 321_7 $?
- Solution: To convert $ 321_7 $ to base ten, expand using powers of 7:
