Four Squared Solution
A B C D
+--+--+--+--+
1 |15|10|1 |8 |
+--+--+--+--+
2 |4 |7 |16|13|
+--+--+--+--+
3 |9 |12|5 |2 |
+--+--+--+--+
4 |6 |3 |14|11|
+--+--+--+--+
The key to the solution lies in comparing (in Grid B) the values of 4C and 3D, which share two
common neighbours (3C and 4D). As there is a difference of ten in their values, 4B is x and
2D is x + 10.
Therefore, the possible values of 4B are 1, 2, 3, 4, 5 and 6; and of 2D, 11, 12, 13, 14, 15 and 16.
However, from 1D, 1C + 2D = 14, therefore, the only possible values for 2D are 11, 12 and 13; and
for 4B, 1, 2 and 3.
From 4A, if 4B is 1, then 3A is 11; but if 4B is 1, then 2D is 11, so, there cannot be a 1 in 4B or
11 in 2D.
From 1D, if 2D is 12, then 1C is 2; but if 2D is 12, then 4B is 2, so, there cannot be a 12 in 2D or
a 2 in 4B.
Therefore, 4B is 3; 2D is 13: 3A is 9 and 1C is 1.
Now, compare the values of 3A and 4B, which share two common neighbours (4A and 3B). As there is a
difference of ten in their values, 2A is y and 4C is y + 10.
Therefore, the possible values of 2A are 2, 4, 5 and 6; and of 4C, 12, 14, 15 and 16.
However, from 4D, 4C + 3D = 16, therefore, the only possible values for 4C are 12 and 14; and for
2A, 2 and 4.
From 1A, if 2A is 2, then 1B is 12; but if 2A is 2, then 4C is 12, so there cannot be a 2 in 2A.
Therefore, 2A is 4; 1B is 10; 4C is 14 and 3D is 2.
The numbers in the neighbouring cells of 2B add up to 42. As we have already placed numbers in two
of these, we can deduce that the numbers in cells 3B and 2C must add up to 28 (42-14). The only
possibilities for these are 16 and 12, or 15 and 13. As we have already placed 13, we are left with
16 and 12. The 12 cannot go in 2C, because number 13 is in 2D.
Therefore, the 12 goes in 3B and the 16 in 2C.
The numbers in the neighbouring cells of 2C add up to 26. As we have already placed numbers in two
of these, we can deduce that the numbers in cells 2B and 3C must add up to 12 (26-14). The only two
possible unused numbers are 7 and 5. The 5 cannot go in 2B, because number 4 is in 2A.
Therefore, the 7 goes in 2B and the 5 in 3C.
Finally
1A is 23 - 7 - 1 = 15
1D is 26 - 16 - 2 = 8
4A is 32 - 14 - 12 = 6
4D is 19 - 3 - 5 = 11
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