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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)

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 Multiple solutions. | Comment 1 of 15
There are several relationships involved. In addition to the given equations, these three, derived from those given, are useful:
B-H=C-I
D-C=H-F
F-B=I-D

That means that the contents of A,B,C,D and G determine all the rest:
H=A+B+C-G-D
F=H+C-D
I=C+H-B
J=A+B+C-F-I
and E is whatever is left over.

Trying all possibilities of A,B,C,D and G is easier than trying all possible permutations.

The following program goes through the possibilities, and then it prints out only those solutions where A>G>J, so that reflections and rotations are not considered as different:

CLS
FOR a = 0 TO 9
taken(a) = 1
FOR b = 0 TO 9
IF taken(b) = 0 THEN
taken(b) = 1
FOR c = 0 TO 9
IF taken(c) = 0 THEN
taken(c) = 1
FOR d = 0 TO 9
IF taken(d) = 0 THEN
taken(d) = 1
FOR g = 0 TO 9
IF taken(g) = 0 THEN
h = a + b + c - g - d
IF h >= 0 AND h <= 9 THEN
IF taken(h) = 0 THEN
taken(h) = 1
f = h + c - d
IF f >= 0 AND f <= 9 THEN
IF taken(f) = 0 THEN
taken(f) = 1
i = c + h - b
IF i >= 0 AND i <= 9 THEN
IF taken(i) = 0 THEN
taken(i) = 1
j = a + b + c - f - i
IF j >= 0 AND j <= 9 THEN
IF taken(j) = 0 THEN
taken(j) = 1
IF g + h + i = a + c + f AND c + f + j = b + d + g THEN
IF a > g AND g > j THEN
FOR e = 0 TO 9
IF taken(e) = 0 THEN EXIT FOR
NEXT
r0 = 5 * (ct \ 5): c0 = 10 * (ct MOD 5)
LOCATE r0 + 1, c0 + 5: PRINT STR\$(a)
LOCATE r0 + 2, c0 + 4: PRINT STR\$(b); STR\$(c)
LOCATE r0 + 3, c0 + 3: PRINT STR\$(d); STR\$(e); STR\$(f)
LOCATE r0 + 4, c0 + 2: PRINT STR\$(g); STR\$(h); STR\$(i); STR\$(j)

ct = ct + 1
END IF
END IF
taken(j) = 0
END IF
END IF
taken(i) = 0
END IF
END IF
taken(f) = 0
END IF
END IF
taken(h) = 0
END IF
END IF
END IF
NEXT
taken(d) = 0
END IF
NEXT
taken(c) = 0
END IF
NEXT
taken(b) = 0
END IF
NEXT
taken(a) = 0
NEXT
PRINT ct

The program finds:
```
5         5         6         7         7

1 7       6 2       0 4       1 3       1 5

6 2 4     1 3 9     5 8 2     2 8 5     6 0 3

4 3 9 0   4 8 4 0   2 3 7 1   5 4 6 0   3 4 8 2

7         8         8         8         8

6 0       0 5       2 3       2 4       2 4

1 4 8     6 2 3     1 4 7     3 0 6     6 1 5

3 9 3 2   3 4 9 1   7 5 6 0   6 5 7 1   1 7 9 0

8         8         8         9         9

2 6       2 6       4 3       0 4       0 6

4 1 7     7 0 4     1 2 9     2 6 5     5 1 3

7 5 9 0   4 5 9 3   7 7 6 0   8 3 7 1   8 2 8 4

9         9         9         9         9

1 3       1 4       2 4       3 0       3 1

2 4 6     3 2 6     3 1 7     1 5 6     0 5 7

6 5 7 0   6 5 8 0   6 6 8 0   4 7 4 2   7 6 4 2

9         9         9         9         9

3 2       3 2       3 4       3 5       5 1

0 4 8     1 4 8     2 0 8     4 0 7     0 2 8

8 6 5 1   6 7 6 0   8 6 7 1   7 6 8 2   8 7 3 4

```

--------
Thus there are 25 basic solutions, not counting rotations and reflections.
 Posted by Charlie on 2003-12-03 10:27:56

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