SOLUTION: Find the smallest positive integer n such that the product of n and 999 does not contain the digit 9.

999*1=999
999*2=1998
999*3=2997
999*4=3996
999*5=4995
999*6=5994
999*7=6993
999*8=7992
999*9=8991

So n cannot be a 1 digit number
since every product of a positive digit and 999
is of the form X99Y, where Y is a positive digit.

Let's see if it can be a 2-digit number:

Then the multiplication would look like this

    999
     AB
   C99D
  E99F 
  GHIJD

B cannot be 0 because the result would only amount to annexing
a 0 onto one of those multiples of 999 by a single digit listed above. 
F cannot be 0 since A cannot be 0.  Now regardless of what positive 
digit F is, we will carry a 1 to the next column which would make I 
be 9.

So n cannot be a 2-digit number.

Let's see if n can be a 3-digit number:

    999
    ABC
   D99E
  F99G
 H99I  
 JKLMNE

C cannot be 0 because that would amount
to annexing a 0 onto a multiple of 999
by a two-digit number and we have already 
ruled those out.

C cannot be 1, for then E would be 9.
So the smallest digit we can try for C is therefore 2, so
then we would have:

    999
    AB2
   1998
  F99G
 H99I  
 JKLMN8

So far so good.  B can't be 0 for that would make G be 0,
and therefore N would be 9.  So the smallest digit we can try
for B is 1. That gives:

    999
    A12
   1998
   999
 H99I  
 JKLM88

So far so good.  The smallest digit we can try
for A is also 1.  That gives:

    999
    112
   1998
   999
  999  
 111888

Eureka! That's it! So the smallest possible value 
for n is n=112.

Edwin