A Trexagon puzzle has 21 triangles arranged in a diamond configuration. Place a digit from 1 to 7 in each triangle to find the unique solution. There are three rules to follow:
- No digit is repeated within a horizontal row or any diagonal in any direction
- Zones of triangle groupings, marked by heavy borders, contain an expression that
specifies the result when applying operators + , - , ×, ÷ to the digits in the zone
- No digit is placed in the shaded triangle
The board contains several zones, that is, triangle groupings marked by heavy borders.
Each zone has a desired value and an operator + , - , ×, or ÷. The digits in each zone will produce the
desired value using the operator. Consider the above board containing
twenty-one triangles, each labeled for reference (in a real puzzle, the triangle labels would not appear). The expressions associated
with each zone determine the digits to place in each triangle. When a zone
contains just a single triangle, the identified digit can simply be inserted.
For example, triangle B5 is the only triangle in its zone so it will
contain a 1.
When a zone contains more than one triangle, the
expression determines the result of applying the operator to its digits. The
three-triangle "◊6" zone, for example, must contain three numbers
that multiply to 6. Since these triangles are all on the same row, these three
digits must be different. The numbers 1 to 3 will eventually be placed in these
three triangles, but we need more information. Consider the two-triangle zone
with the expression "◊14". These two digits must multiply to 14 and
are therefore 2 and 7. Because a 2 digit must exist in either triangle B2 or
C1 you can eliminate the 2 digit from triangles D1 and D2 in
the "◊6" zone. Therefore a 2 digit can safely be placed in triangle D3.
In the "+17" zone, the digits must be 7, 6, and 4. Because a 7 digit
must appear in the "◊14" zone, a 7 digit can only be placed in
triangle A2. Because of the 1 in triangle B5 you can put a 1 in
triangle D1 and therefore a 3 in triangle D2. The solution so far
The "◊5" zone must contain a 1 and a 5, but
because of the 1 in B5 you must place the 1 in triangle A3 and a
5 in triangle B3. One can also apply logic to find where a digit is to
be placed. There are seven triangles in the left-most diagonal, from A1 in
the upper left down to D2 in the lower right; therefore all 7 digits must
appear in these triangles. Where does a 5 belong? It canít be in the "◊14"
zone, nor can it be in the "+17" zone (because then both A1 and
B1 would contain a 5, which is impossible). So the only place for a 5 in
this long diagonal is triangle C2.
The top row from A1 to A7 must contain a 5
digit, and it can only exist in A7 given the 5 placed in triangles B3 and C2.
It is possible to eliminate a digit from consideration in a
triangle. Because of the placement of 1 in B5 there can be no 1 in the "+9"
zone; therefore, this zone must contain the digits 2, 3, 4 (which is the only
remaining way for three different digits to sum to 9). Because of the other
existing digits in the top row (a 7 in A2, a 1 in A3, and a 5 in A7),
this means the only remaining digit left for A6 is 6. You can now place
a 7 in B6. Look at the remaining "+18" zone and you can see
that a 4 and 6 must be placed (in some arrangement) in C4 and C5.
Even without knowing the proper arrangement, you can add 4+6+5=15 and determine
that a 3 must be placed in C3. The solution so far is:
Recall that the "+9" zone must contain a 2, 3,
and 4; because of the 3 in C3, you must place a 3 in A4. A 2 must
be placed in the top row (from A1 to A7), but it canít be placed
in A1 because then the "+17" zone would have no solution; thus
place a 2 in A5, a 4 in A1, and a 6 in B1. Complete the "+9" zone
by placing a 4 in B4.
Finally, the digits 4 and 6 need to be placed in C4 and C5.
However, with the 4 in B4 you must place a 4 in C4 and a 6 in C5.
To finish the puzzle, observe that the 7 in B6 means you must place a 7
in C1 and finally a 2 in B2.
You can confirm the solution is correct, because all digits in every row and diagonal
in any direction are different.