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ramada plaza by wyndham marco polo beach resort - Redraw

📅 May 5, 2026 👤 scraface

May 05, 2026

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📌 a^3 + b^3 = 7^3 - 3(10)(7) = 343 - 210 = 133

📌 Thus, the value is $ oxed{133} $.Question: How many lattice points lie on the hyperbola $ x^2 - y^2 = 2025 $?

📌 Solution: The equation $ x^2 - y^2 = 2025 $ factors as $ (x - y)(x + y) = 2025 $. Since $ x $ and $ y $ are integers, both $ x - y $ and $ x + y $ must be integers. Let $ a = x - y $ and $ b = x + y $, so $ ab = 2025 $. Then $ x = rac{a + b}{2} $ and $ y = rac{b - a}{2} $. For $ x $ and $ y $ to be integers, $ a + b $ and $ b - a $ must both be even, meaning $ a $ and $ b $ must have the same parity. Since $ 2025 = 3^4 \cdot 5^2 $, it has $ (4+1)(2+1) = 15 $ positive divisors. Each pair $ (a, b) $ such that $ ab = 2025 $ gives a solution, but only those with $ a $ and $ b $ of the same parity are valid. Since 2025 is odd, all its divisors are odd, so $ a $ and $ b $ are both odd, ensuring $ x $ and $ y $ are integers. Each positive divisor pair $ (a, b) $ with $ a \leq b $ gives a unique solution, and since 2025 is a perfect square, there is one square root pair. There are 15 positive divisors, so 15 such factorizations, but only those with $ a \leq b $ are distinct under sign and order. Considering both positive and negative factor pairs, each valid $ (a,b) $ with $ a

📌 e b $ contributes 4 lattice points (due to sign combinations), and symmetric pairs contribute similarly. But since $ a $ and $ b $ must both be odd (always true), and $ ab = 2025 $, we count all ordered pairs $ (a,b) $ with $ ab = 2025 $. There are 15 positive divisors, so 15 positive factor pairs $ (a,b) $, and 15 negative ones $ (-a,-b) $. Each gives integer $ x, y $. So total 30 pairs. Each pair yields a unique lattice point. Thus, there are $ oxed{30} $ lattice points on the hyperbola.

📌 Question: What is the remainder when $ 12003 + 12005 + 12007 + 12009 $ is divided by $ 16 $?

📌 Solution: Compute the sum: $ 12003 + 12005 + 12007 + 12009 $. This is an arithmetic sequence with first term 12003, last term 12009, 4 terms. Sum = $ rac{4}{2} (12003 + 12009) = 2 \cdot 24012 = 48024 $. Now compute $ 48024 \mod 16 $. Note that $ 10000 \equiv 0 \pmod{16} $, so we only need the last 4 digits: $ 8024 $. Now $ 8024 \div 16 $: $ 16 imes 500 = 8000 $, $ 8024 - 8000 = 24 $, $ 24 \div 16 = 1 $

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