求解4779四个数字如何不重复组成24

发布网友发布时间：2024-10-24 00:13

共1个回答

热心网友时间：2024-11-10 22:00

4779
在只有加减乘除的情况下，无法得到24
允许开根号时
根号9=3
那么变成 4773
(4*7)-(7-3)
((4*7)-7)+3
((4*7)+3)-7
(4*7)+(3-7)
(4*7)-(7-3)
((4*7)-7)+3
((4*7)+3)-7
(4*7)+(3-7)
7-(4-(7*3))
(7-4)+(7*3)
((7-4)*7)+3
(7*4)-(7-3)
((7*4)-7)+3
7-(4-(3*7))
(7-4)+(3*7)
((7*4)+3)-7
(7*4)+(3-7)
(7*(7-4))+3
7+((7*3)-4)
(7+(7*3))-4
(7*(7-3))-4
(7*3)-(4-7)
((7*3)-4)+7
((7-3)*7)-4
7+((3*7)-4)
(7+(3*7))-4
((7*3)+7)-4
(7*3)+(7-4)
7-(4-(7*3))
(7-4)+(7*3)
((7-4)*7)+3
(7*4)-(7-3)
((7*4)-7)+3
7-(4-(3*7))
(7-4)+(3*7)
((7*4)+3)-7
(7*4)+(3-7)
(7*(7-4))+3
7+((7*3)-4)
(7+(7*3))-4
(7*(7-3))-4
(7*3)-(4-7)
((7*3)-4)+7
((7-3)*7)-4
7+((3*7)-4)
(7+(3*7))-4
((7*3)+7)-4
(7*3)+(7-4)
3-((4-7)*7)
3+((4*7)-7)
(3+(4*7))-7
3-((4-7)*7)
3+((4*7)-7)
(3+(4*7))-7
3-(7-(4*7))
(3-7)+(4*7)
3-(7*(4-7))
3+((7-4)*7)
3+((7*4)-7)
(3+(7*4))-7
(3*7)-(4-7)
((3*7)-4)+7
3-(7-(7*4))
(3-7)+(7*4)
3+(7*(7-4))
((3*7)+7)-4
(3*7)+(7-4)
3-(7-(4*7))
(3-7)+(4*7)
3-(7*(4-7))
3+((7-4)*7)
3+((7*4)-7)
(3+(7*4))-7
(3*7)-(4-7)
((3*7)-4)+7
3-(7-(7*4))
(3-7)+(7*4)
3+(7*(7-4))
((3*7)+7)-4
(3*7)+(7-4)
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