The divisors of 2023 are \(1, 7, 17, 49, 119, 289, 2023\). The largest divisor is 2023 itself. If \(d = 2023\), then \(x + y = 1\), which implies one of \(x\) or \(y\) is 0, contradicting \(x, y\) being positive integers. So, the largest feasible \(d\) is 289.

Sun, 19 Apr 2026 03:23:01 +0000

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