Each letter stands for a different number. We already know that D = 6 and the total for each row and column.
Can you find the values of A, B, C, E, F, G. H, J?
H | D | B | A | H | A | J | E | J | = 55 |
F | E | B | B | H | C | G | E | G | = 68 |
C | G | F | B | J | C | G | B | G | = 83 |
C | G | J | H | G | G | H | C | H | = 88 |
A | J | G | H | J | F | A | G | B | = 81 |
B | A | G | G | B | F | C | G | B | = 83 |
D | A | H | G | B | J | C | H | C | = 70 |
F | B | H | J | A | D | D | D | A | = 65 |
F | C | F | J | C | H | E | A | E | = 70 |
80 | 76 | 76 | 67 | 61 | 85 | 79 | 71 | 68 |
|
There is more than one way to solve this puzzle. Below is one of the ways.
C stands for Column in C1,C2…..C9; and R stands for Row in R1,R2…..R9
1) from C4&C5, get G>C = 6
2) from C6&C7, get FF>EG = 6, from 1) get FF>EC = 12
3) from C3&C4, get FF>AJ = 9, from 2) get EG>AJ = 3 and AJ>EC = 3
4) from C7&C8, get CJ>BE = 8, then ACJ>ABE = 8. from 3) AJ>EC = 3, so that ECC>ABE = 5, CC>AB = 5
5) from R3&R6, get CJ = AB, from 4) CC>AB = 5, therefore A>E = 8, and C>J = 5, G>J = 11,
6) from C7&R7, get EG>BH = 9, from 2) FF>EG = 6, get FF>BH = 15 and from 3) FF>AJ = 9, get AJ>BH = 6
7) from R4&R7, get GGH>ABD = 18, from 4) CC>AB = 5, get GGH>CCD = 13, and from 1) get H>D = 1
8) from C2&C4, get ACDE>BHHJ = 9 and from 7) H>D = 1, so that ACDE>BDDJ = 11, from 3) AJ>EC = 3, get AADJ>BDDJ = 14, AA>BD = 14, from 7) H>D = 1, get AA>BH = 13, from 3 & 6) EG>AJ = 3 and EG>BH = 9, so that AJ>BH = 6 there for A>J = 7, J>E = 1 and from 5) CJ = AB so that C>B = 7, and from 3) FF>AJ = 9 so that FF>BH = 15, therefor F>A = 1, F>J = 8, F>C
作者: husonghu 时间: 2010-1-3 20:47
标题: 回复:The Answer for the Brainteaser-4 [:)]
作者: GiveItATry 时间: 2010-1-4 23:19
标题: 回复:回复:The Answer for the Brainteaser-4 [:)]
欢迎光临 珍珠湾ART (http://art.zhenzhubay.com/)
Powered by Discuz! X3