Question 851668
I have before me three numeric palindromes (numbers which read the same backwards and forwards, like 838).  The first number is two digits long; the second is three digits, and when we add those two numbers together we get the third number, which is four digits long.  What are the three numbers?
<pre>

Rich beat me to your problem, but I'll bet you'll like my way of 
explaining it better:

I will use capital letters A,B,C,D,E for the digits.
The numbers above with arrows pointing down are to
label the columns of digits, column 1, column 2,
column 3, and column 4:

1234
&#8595;&#8595;&#8595;&#8595;

  AA
<u>+BCB</u>
DEED

The only way we could have a D in column 1 is for there
to have been 1 to carry from column 2.  So we have D=1, 
so replacing D by 1, we have

1234
&#8595;&#8595;&#8595;&#8595;

  AA
<u>+BCB</u>
1EE1

To have gotten a 1 in column 4,  it could only be that
A+B=11. So there was a 1 'to carry' from column 4 to 
column 3. So we put a 1 above column 3

1234
&#8595;&#8595;&#8595;&#8595;
  <sub>1</sub>
  AA
<u>+BCB</u>
1EE1

Even if A and C were as large as they could be (both 9's) there
could be at most 1 'to carry' from column 3 to column 2, since
if A and C were both 9's we could only get 1+A+C=1+9+9=19, so
the 'carry' from column 3 to column 2 could not be as much as 2.
And we have already determined that there must be a carry from column 
3 to column 2, so the 'carry' to column 2 must be 1.  So we
put a 1 above column 2:

1234
&#8595;&#8595;&#8595;&#8595;
 <sub>1 1</sub>
  AA
<u>+BCB</u>
1EE1

The only digit that B could be is 9, because that is the only
digit that a 'carry' of 1 to results in a 2-digit number.
And that 2 digit number is, of course 10.  So we know that 
B=9 and E=0, so we replace those letters by digits:

1234
&#8595;&#8595;&#8595;&#8595;
 <sub>1 1</sub>
  AA
<u>+9C9</u>
1001

Column 4 tells us that A=2. So we have

1234
&#8595;&#8595;&#8595;&#8595;
 <sub>1 1</sub>
  22
<u>+9C9</u>
1001      

Now C has to cause column 3 to add up to 10,
So C=7, and the addition is
 <sub>1 1</sub>
  22
<u>+979</u>
1001

Edwin</pre>