有一九宫格数学题横竖都等于四给了一个提示是九

2025-03-15 00:07:20
推荐回答(5个)
回答1:

首先这个如果是九宫格的话,数字应该是1/2/3/4/5/6/7/8/9这几个数字,题中已经出现了9,我们开始分析,A+B=13,也就是5+8或者6+7,看第一列得知A+C/F=4,所有数都是正整数,而上述数字最小的正整数商是2,也就是说A要小于等于2,这和A是5/6/7/8中的一个矛盾,所以显然不是九宫算法了,那么假如所有数都是不重复正整数,由A+C/F=4得知,因为C/F的值是2、3、4、5……的整数列,所以A的值取整那就1/2中的一个,假设A=1,那B=12,已知条件E+H=5,那么E和H只能是2和3,再设E为2,H为3,由已知条件F+G=7,得知1+6或2+5或3+4才能满足,但是要不重复,所以E=2,H=3不成立,同理可知E=3,H=2也不成立,所以A不等于1,那么A只能等于2,那B=11,同理假设可知无正确答案,那么我们把0包含进去,同理推理得知A=2,B=11,C=4,D=1,E=0,F=2,G=7,H=5

回答2:

5 + 8 — 9 =4
+ — —
7 — 6 x 4 =4
÷ x —
3 + 2 — 1 =4
= = =
4 4 4此题不能先乘除后加减,要自左至右自上至下运算,空格只能填1至8每数一次。

回答3:

9+4-9=4
+ - -
7-3 × 1=4
÷ × -
4+4-4=4
= = =
4 4 4

回答4:

9-4-9=4
+ - -
3-2-1=4
/ - -
3+2-1=4
= = =
4 4 4

回答5:

49
814
051

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