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treereding 寫到:下列方程組 {𝑥 + 𝑦 = 3(𝑧 + 𝑢)
                  𝑥 + 𝑧 = 5(𝑦 + 𝑢)
                  𝑥 + 𝑢 = 7(𝑦 + 𝑧)
的解 (𝑥, 𝑦, 𝑧, 𝑢)，其中 𝑥、𝑦、𝑧、𝑢 皆為正整數，求 𝑥可能的最小值為何？

y - z = 3(z+u) - 5(y+u) → 0 = 4z-6y-2u → 2z = 3y + u
z - u = 5(y+u) - 7(y+z) → 0 = 6u-2y-3z → 3z = 6u - 2y
→y = 9/13 u, z = 20/13 u
→ u = 13m, y = 9m, z = 20m

x  = 3(z + u) - y = 99/13 u = 99m

minimu x = 99 (when m = 1)

感恩大大!!

treereding 寫到:下列方程組 {𝑥 + 𝑦 = 3(𝑧 + 𝑢)
                  𝑥 + 𝑧 = 5(𝑦 + 𝑢)
                  𝑥 + 𝑢 = 7(𝑦 + 𝑧)
的解 (𝑥, 𝑦, 𝑧, 𝑢)，其中 𝑥、𝑦、𝑧、𝑢 皆為正整數，求 𝑥可能的最小值為何？

y - z = 3(z+u) - 5(y+u) → 0 = 4z-6y-2u → 2z = 3y + u
z - u = 5(y+u) - 7(y+z) → 0 = 6u-2y-3z → 3z = 6u - 2y
→y = 9/13 u, z = 20/13 u
→ u = 13m, y = 9m, z = 20m

x  = 3(z + u) - y = 99/13 u = 99m

minimu x = 99 (when m = 1)

下列方程組 {𝑥 + 𝑦 = 3(𝑧 + 𝑢)
                  𝑥 + 𝑧 = 5(𝑦 + 𝑢)
                  𝑥 + 𝑢 = 7(𝑦 + 𝑧)
的解 (𝑥, 𝑦, 𝑧, 𝑢)，其中 𝑥、𝑦、𝑧、𝑢 皆為正整數，求 𝑥可能的最小值為何？