<PRE><B>Question 1209560</B>
&lt;br&gt;
I will assume the cryptarithm is this:&lt;br&gt;&lt;pre&gt;

  ABC
   DE
 + BE
  ---
  BDC&lt;/pre&gt;

This is not a very well-made cryptarithm; there are many solutions.&lt;br&gt;
(1) In the units column, C+E+E yields final digit C.  That means E is either 0 or 5.&lt;br&gt;
(2) In the tens column, B+D+B, plus possibly a carry of 1, yields digit D in the sum, so B is also either 0 or 5.&lt;br&gt;
(3) Since in the tens column B+D+B yields digit D in the sum, there can not be a carry of 1 from the units column.&lt;br&gt;
(4) From (1), (2), and (3), E must be 0 and B must be 5.&lt;br&gt;
(5) In the hundreds column, A plus a carry of 1 yields leading digit B (=5) in the sum, so A is 4.&lt;br&gt;
So we know the addition looks like this:&lt;br&gt;&lt;pre&gt;

  45C
   D0
 + 50
  ---
  5DC&lt;br&gt;&lt;/pre&gt;
There are no further restrictions, so C and D can be any digits.  The three digits 0, 4, and 5 have been used; assuming the letters represent different digits, there are 7 choices remaining for C and then 6 choices remaining for D, making a total of 7*6 = 42 solutions.&lt;br&gt;
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