<PRE><B>Question 1204315</B>
&lt;br&gt;
Although a formal algebraic solution is good, and the student should be able to understand it, a very different and much easier path to the answer is possible.&lt;br&gt;
Given that the 3-digit number ABC divided by the 2-digit number AC gives a quotient of 11 with no remainder, we know that AC*11 = ABC.  So look at that multiplication in the way we learn multiplication in grade school.&lt;br&gt;&lt;pre&gt;

     A C
   X 1 1
   ------
     A C
   A C
  -------
   A B C&lt;/pre&gt;
The problem asks for the largest possible value of the 3-digit number ABC.  So let&#39;s see if the condition can be satisfied if A is 9.&lt;br&gt;&lt;pre&gt;

     9 C
   X 1 1
   ------
     9 C
   9 C
  -------
   9 B C&lt;/pre&gt;
We can see that with A = 9, C must be 0, giving us&lt;br&gt;&lt;pre&gt;

     9 0
   X 1 1
   ------
     9 0
   9 0
  -------
   9 9 0&lt;/pre&gt;
So A = B = 9 and C = 0.  Generally, in problems like this, it is specified that different letters represent different digits.  However, that is not specified in the statement of this problem.  So the answer could be&lt;br&gt;
ANSWER: ABC = 990&lt;br&gt;
Assuming any information that is not given in a problem is never good mathematics.  However, if we assume that the letters represent different digits, then A = 9 doesn&#39;t work.  So again looking for the largest possible value of the 3-digit number ABC, we try A = 8:&lt;br&gt;&lt;pre&gt;

     8 C
   X 1 1
   ------
     8 C
   8 C
  -------
   8 B C&lt;/pre&gt;
Here we can see that B can be at most 9, which means C can be at most 1.  And since we want ABC to be the largest possible, we choose C = 1, giving us&lt;br&gt;&lt;pre&gt;

     8 1
   X 1 1
   ------
     8 1
   8 1
  -------
   8 9 1&lt;/pre&gt;&lt;br&gt;
And in this case we have&lt;br&gt;
ANSWER: ABC = 891&lt;br&gt;
----------------------------------------------------------------------&lt;br&gt;
And then here is another solution VERY different from the others, and MUCH faster.&lt;br&gt;
Using the divisibility rule for 11 in the problem, we know immediately that&lt;br&gt;
A + C = B&lt;br&gt;
Then, knowing that A, B, and C are single digit integers, we immediately know that B is 9 and A + C is 9.&lt;br&gt;
Then, knowing that we want ABC to be as large as possible, we quickly find ABC = 990 if A and B can be the same digit, or ABC = 891 if they can&#39;t be the same.&lt;br&gt;

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