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mississippi state football coach

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Maximizing \(d\) requires minimizing \(m + n\), with \(m, n \geq 1\) and \(\gcd(m, n) = 1\). The minimum value of \(m + n\) is 2, achieved when \(m = n = 1\), which are coprime. Then:
d = \frac{2025}{2} = 1012.5
But \(d\) must be an integer. So \(m + n\) must divide 2025 evenly, and be at least 2. We seek the largest divisor \(d\) of 2025 such that \( \frac{2025}{d} \geq 2 $ and $m + n = \frac{2025}{d}$ is minimized among valid coprime pairs. The smallest possible $m + n \geq 2$ is 3 (e.g., $m = 1, n = 2$), which is coprime. Then:

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