First, we claim that
is the
digit number with all digits equal to one.
Note that
is a one followed by
zeroes, so subtracting one gives
nines. Divide that number by nine and you get
ones, completing the proof.
Therefore, we have that
,
, and
.
Let
. Then, we have:
.
Multiplying by
gives:
Now, note that
is not a valid input, since
requires
, so we safely divide by
to get:
Because this is now a linear equation in
, it has either zero, one, or infinitely many solutions. Obviously, we need the latter to occur, which happens when
and
, since the coefficient of
must cancel to zero and thus the RHS must equal zero as well.
Since
, we must have
. Since
must be a multiple of three, we plug in values. If
, we get
and thus
, which is impossible. So
doesn't work.
With
,
, and thus
, so
, so
. This gives
as one possibility.
With
,
, and thus
, so
. This obviously is worse.
We've gone through all the cases and the two possibilities are
and
, so our answer is
.
