```So far the question is lame as a question. (Barak (bap@valaga.salk.edu)
has pointed out.)

Of course, we want to neglect the trivial answer, but I should give
you more restriction for the question. If not, the answer could be
arbitrarily given.

The restriction is:

The domain of the ilog function (thus {1,2,3,...,N}) must satisfy
usual multiplication rule if the result does not exceed the maximum
number N.

For example, let us think the case N=4:

The multiplication table must satisfy at least,

1 x 1 = 1    1 x 2 = 2       1 x 3 = 3    1 x 4 = 4
2 x 1 = 2       3 x 1 = 3    4 x 1 = 4
2 x 2 = 4

Therefore the part of the multiplication table is
predetermined in the question from the first as

+---++---+---+---+---+
| X || 1 | 2 | 3 | 4 |
+===++===+===+===+===+
| 1 || 1 | 2 | 3 | 4 |
+---++---+---+---+---+
| 2 || 2 | 4 | * | * |
+---++---+---+---+---+
| 3 || 3 | * | * | * |
+---++---+---+---+---+
| 4 || 4 | * | * | * |
+---++---+---+---+---+

Since 2 x 3 (or other cases expressed by "*") comes to
larger than N(=4) under the conventional rule, the multiplication
rule is determined by the demand of ilog() conditions.

Therefore, the question is to give the part of "*" in
the above table such that ilog() conditions (as follows)
are satisfied.

ilog(2 x 2) = ilog(2) + ilog(2).
ilog(1 x 2) = ilog(1) + ilog(2).
....

Obviously, there still are  multiple answers  (but not so many)
in the above example. The question is whether or not we can
always obtain multiple answers for the question in the arbitrary N.

```