Problem 3
a) Label the grid with top leftmost square (1, 1), and the x coordinate progressing as we move right,
the y coordinate progressing as we move down.
Note that on the diagonal where the x coordinate is equal to the y coordinate, not every possible
entry from 1 to 2 · 1997 − 1 can appear. This is because there are exactly 1997 values on that
diagonal, while there are 2 · 1997 − 1 > 1997 values in the possible entries.
Now, consider a value k such that k does not appear on the diagonal.
Note that in a single appearance of k, it must have coordinates (i, j), where i ̸= j. Then, note that
no other appearance of k can have neither i nor j in its coordinates.
Then, note that each appearance of k eliminates two possible values from 1, 2, . . . , 1997. However,
all of these values must appear in a pair of coordinates of a point labelled k. Since 1997 is odd, this
is a contradiction.
b) We provide a construction for all n = 2k .
First, we put 1 on the whole of the diagonal. This satisfies that there is exactly one of 1 in ith row
and ith column.
We will construct 2k+1 − 2 good sets by considering the difference between the coordinates in each
of the sets. For example, if the difference is 1, we can create a good set with coordinates (1, 2),
(3, 4), ..., where each pair of coordinates has y − x = 1, where y and x are the y and x coordinate
respectively.
Let this difference be d.
We move into cases:
Case 1: If d is odd.
Then note that d and 2k are coprime.
Consider the coordinate set (d, 2d), (3d, 4d), . . . , ((2k − 1) · d, 2k · d). Note that each of these are
distinct, since id ̸≡ jd (mod 2k ) if i ̸≡ j (mod 2k ). So, this set is good.
Also consider the coordinate set (2d, 3d), (4d, 5d), . . . , (2k d, d). This is good for the same reasons as
before.
Thus, each d in this case contributes 2 good sets.
Case 2: If d = c2e , where e < k − 1.
Another good subset is (d + 1, 2d + 1), (3d + 1, 4d + 1), . . . , (2k−e d + 1) and (d + 2, 2d + 2), (3d +
2, 4d + 2), . . . , (2k−e d + 1), and so on. This is good for the same reasons as described above.
Thus, each d in this case contributes 2 good sets.
Case 3: If d = 2k−1 .
Consider the set (1, 1 + 2k−1 ), (2, 2 + 2k−1 ), . . . , (2k−1 − 1, 2k−1 − 1 + 2k−1 ). This is clearly good
since all of the values of different coordinates are different (mod 2)k−1 , and it covers all values
(mod 2k ).
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