Please help with number 27 thanks
2 answers:
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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 
.
Note that
Therefore, we have that
Let
Multiplying by
Now, note that
Because this is now a linear equation in
Since
With
With
We've gone through all the cases and the two possibilities are