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一元一次方程式 ~ 急
Dec 31st 2013, 10:55

設原數為 ab8c
則逆順序為 c8ba
c8ba = 9 * ab8c
1000c+800+10b+a = 9(1000a+100b+80+c)
1000c+800+10b+a = 9000a+900b+720+9c  
等式左邊之個位數字為a, 等式右邊之個位數字為9c除以10的餘數.
a = 9c mod 10 .....(1)

890b = -8999a+991c+80
b = (-8999a+991c+80) / 890 .....(2)

利用(1),(2)可計算
c,a,b之值為:
檢視圖片  
唯一能成立之解為:
c = 9 , a = 1 , b = 0
故原數為1089

Ans: 1089

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