小学奥数题求解：用1—9个数组成三道加法算式，每个数字只能用一次。

发布网友 发布时间：2022-04-28 18:23

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热心网友 时间：2023-09-14 04:33

这是不可能的，九个数中有4个偶数，偶加偶还为偶，偶加奇为奇。三个等式中不可能都为偶加奇为奇，不然会剩下一个偶数。所以会有一个是偶加偶等于偶，这样就只剩下一个偶数了，你让它加奇怎么能等于奇数啊

热心网友 时间：2023-09-14 04:33

刚子好聪明，果然无解！只要先运算几步就发现这道题最后一组总是少一。厉害！

热心网友 时间：2023-09-14 04:33

Sub 九宫格()
Dim A1%, A2%, A3%
Dim B1%, B2%, B3%
Dim C1%, C2%, C3%
Dim I%, J%, K%
Dim X%
X = 0
For I = 102 To 987
A1 = Int(I / 100)
A2 = Int((I Mod 100) / 10)
A3 = I - A1 * 100 - A2 * 10
If A1 <> A2 And A1 <> A3 And A2 <> A3 And A1 <> X And A2 <> X And A3 <> X Then
For J = 102 To 987
B1 = Int(J / 100)
B2 = Int((J Mod 100) / 10)
B3 = J - B1 * 100 - B2 * 10
If B1 <> B2 And B1 <> B3 And B2 <> B3 And B1 <> X And B2 <> X And B3 <> X Then
If A1 <> B1 And A1 <> B2 And A1 <> B3 And _
A2 <> B1 And A2 <> B2 And A2 <> B3 And _
A3 <> B1 And A3 <> B2 And A3 <> B3 Then
For K = 102 To 987
C1 = Int(K / 100)
C2 = Int((K Mod 100) / 10)
C3 = K - C1 * 100 - C2 * 10
If C1 <> C2 And C1 <> C3 And C2 <> C3 And C1 <> X And C2 <> X And C3 <> X Then
If I <> J And I <> K Or J <> K Then
If A1 <> C1 And A1 <> C2 And A1 <> C3 And _
A2 <> C1 And A2 <> C2 And A2 <> C3 And _
A3 <> C1 And A3 <> C2 And A3 <> C3 And _
B1 <> C1 And B1 <> C2 And B1 <> C3 And _
B2 <> C1 And B2 <> C2 And B2 <> C3 And _
B3 <> C1 And B3 <> C2 And B3 <> C3 Then
If A1 + B1 = C1 And A2 + B2 = C2 And A3 + C3 = B3 Then
Debug.Print I & Chr(10) & J & Chr(10) & K & Chr(10)
End If
End If
End If
End If
Next K
End If
End If
Next J

End If
Next I
End Sub
没有答案，计算没有结果

热心网友 时间：2023-09-14 04:34

1+6+8=3+5+7=2+9+4

热心网友 时间：2023-09-14 04:35

182+457=639