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 Tricky Triangle (Posted on 2003-12-03)
Replace the letters in the diagram with a different number from 0 to 9, such that the sum of the four numbers on all edges are the same, and the sum of the three numbers on all three corners are the same.
```          A
B   C
D   E   F
G   H   I   J
```

In other words, A+C+F+J = A+B+D+G = G+H+I+J and
A+B+C = D+G+H = F+I+J.

 No Solution Yet Submitted by Kelsey Rating: 3.2500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Full solution | Comment 3 of 15 |
There are a dozen solutions:

ABCDEFGHIJ
0594378261
0957341628
1574639028
1672398450
1753648209
1769320548
8230679451
8246351790
8327601549
8425360971
9042658371
9405621738

They were produced by the following awk program:
```
BEGIN {

for (A=0; A<=9; A++)

{ for (B=0; B<=9; B++) if (B!=A)

{ for (C=0; C<=9; C++) if ((C!=A) && (C!=B))

{ for (D=0; D<=9; D++) if ((D!=A) && (D!=B) && (D!=C))

{ for (E=0; E<=9; E++) if ((E!=A) && (E!=B) && (E!=C) && (E!=D))

{ for (F=0; F<=9; F++) if ((F!=A) && (F!=B) && (F!=C) && (F!=D) && (F!=E))

{ for (G=0; G<=9; G++) if ((G!=A) && (G!=B) && (G!=C) && (G!=D) && (G!=E) && (G!=F))

{ for (H=0; H<=9; H++) if ((H!=A) && (H!=B) && (H!=C) && (H!=D) && (H!=E) && (H!=F) && (H!=G))

{ for (I=0; I<=9; I++) if ((I!=A) && (I!=B) && (I!=C) && (I!=D) && (I!=E) && (I!=F) && (I!=G) && (I!=H))

{ J = 45 - (A+B+C+D+E+F+G+H+I)

if ((J!=A) && (J!=B) && (J!=C) && (J!=D) && (J!=E) && (J!=F) && (J!=G) && (J!=H) && (J!=I))

{ if ((A+B+D+G==G+H+I+J) && (A+B+D+G==A+C+F+J) && (A+B+C==D+G+H) && (D+G+H==F+I+J))

{ print A B C D E F G H I J

}

}

}

}

}

}

}

}

}

}

}

}

Edited on December 3, 2003, 11:25 am

Posted by Federico Kereki
on 2003-12-03 10:54:12

```

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