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ybor city tampa fl 33605 - NBX Soluciones

📅 May 5, 2026 👤 scraface

May 05, 2026

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📌 We aim to maximize one of \(6a\), \(6b\), or \(6c\). Without loss of generality, let’s maximize \(6a\). This implies maximizing \(a\) subject to \(a + b + c = 14\) and \(\gcd(a, b, c) = 1\).

📌 Start by examining possible values for \(a\):

📌 If \(a = 12\), then \(b + c = 2\). Possible pairs \((b, c)\) are \((1, 1)\), but \(\gcd(12, 1, 1) = 1\), which is valid.

📌 Thus, \(a = 12\) is feasible, giving the maximum erosion rate for one region as:

📌 6a = 6 \times 12 = 72

📌 Therefore, the maximum possible erosion rate for any one region is \(\boxed{72}\).

📌 A hydrologist is modeling water flow through a system of aquifers. If the flow rates through three distinct aquifers are three consecutive integers, and their sum is 90 cubic meters per hour, what is the flow rate through the aquifer with the highest rate?

📌 Let the flow rates through the three aquifers be \(x-1\), \(x\), and \(x+1\). The sum of these rates is:

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Solution: We compute the sum $ \sum_{k=1}^{24} 3^k \mod 7 $. This is a...
\sum_{k=1}^{24} 3^k = 3 \cdot \frac{3^{24} - 1}{3 - 1} = \frac{3}{2} (...
But working modulo $7$, we can compute the sum directly using cyclicit...
3^1 &\equiv 3 \mod 7 \\
3^2 &\equiv 9 \equiv 2 \mod 7 \\
3^3 &\equiv 3 \cdot 2 = 6 \mod 7 \\
3^4 &\equiv 3 \cdot 6 = 18 \equiv 4 \mod 7 \\
3^5 &\equiv 3 \cdot 4 = 12 \equiv 5 \mod 7 \\
3^6 &\equiv 3 \cdot 5 = 15 \equiv 1 \mod 7 \\
So the powers of $3 \mod 7$ repeat every 6 terms. The cycle length is...

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