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Home » \(a_5 = 3(5)^2 + 2(5) + 1\). - Treasure Valley Movers

\(a_5 = 3(5)^2 + 2(5) + 1\). - Treasure Valley Movers

📅 April 16, 2026 👤 scraface

Apr 16, 2026

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📌 The area of the original square is \(64\) square units, so each side length \(s\) is:

📌 s = \sqrt{64} = 8 \text{ units}

📌 Increasing each side by \(50\%\) gives a new side length:

📌 s_{\text{new}} = 8 + 0.5 \times 8 = 1.5 \times 8 = 12 \text{ units}

📌 The new area is:

📌 A_{\text{new}} = (12)^2 = 144 \text{ square units}

📌 Alternatively, since area scales by the square of the side length factor (\(1.5^2 = 2.25\)):

📌 A_{\text{new}} = 64 \times 2.25 = 144

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\Rightarrow 55^2 \equiv 6^2 = 36 \equiv 1 \pmod{7}
Alternatively, $ 3025 \div 7 = 432 \text{ remainder } 1 $, so $ 3025 \...
Therefore, the remainder is $ \boxed{1} $.
Question: What is the greatest common divisor of $ 2^{12} - 1 $ and $...
Solution: We use the identity:
\gcd(2^a - 1, 2^b - 1) = 2^{\gcd(a,b)} - 1
Here, $ a = 12 $, $ b = 18 $, so $ \gcd(12, 18) = 6 $.
Thus,
\gcd(2^{12} - 1, 2^{18} - 1) = 2^6 - 1 = 64 - 1 = 63
Therefore, the greatest common divisor is $ \boxed{63} $.

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