document.write( "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? \n" ); document.write( "
Algebra.Com's Answer #512930 by Edwin McCravy(20056)\"\" \"About 
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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?
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document.write( "Rich beat me to your problem, but I'll bet you'll like my way of \r\n" );
document.write( "explaining it better:\r\n" );
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document.write( "I will use capital letters A,B,C,D,E for the digits.\r\n" );
document.write( "The numbers above with arrows pointing down are to\r\n" );
document.write( "label the columns of digits, column 1, column 2,\r\n" );
document.write( "column 3, and column 4:\r\n" );
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document.write( "1234\r\n" );
document.write( "↓↓↓↓\r\n" );
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document.write( "  AA\r\n" );
document.write( "+BCB\r\n" );
document.write( "DEED\r\n" );
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document.write( "The only way we could have a D in column 1 is for there\r\n" );
document.write( "to have been 1 to carry from column 2.  So we have D=1, \r\n" );
document.write( "so replacing D by 1, we have\r\n" );
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document.write( "1234\r\n" );
document.write( "↓↓↓↓\r\n" );
document.write( "\r\n" );
document.write( "  AA\r\n" );
document.write( "+BCB\r\n" );
document.write( "1EE1\r\n" );
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document.write( "To have gotten a 1 in column 4,  it could only be that\r\n" );
document.write( "A+B=11. So there was a 1 'to carry' from column 4 to \r\n" );
document.write( "column 3. So we put a 1 above column 3\r\n" );
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document.write( "1234\r\n" );
document.write( "↓↓↓↓\r\n" );
document.write( "  1\r\n" );
document.write( "  AA\r\n" );
document.write( "+BCB\r\n" );
document.write( "1EE1\r\n" );
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document.write( "Even if A and C were as large as they could be (both 9's) there\r\n" );
document.write( "could be at most 1 'to carry' from column 3 to column 2, since\r\n" );
document.write( "if A and C were both 9's we could only get 1+A+C=1+9+9=19, so\r\n" );
document.write( "the 'carry' from column 3 to column 2 could not be as much as 2.\r\n" );
document.write( "And we have already determined that there must be a carry from column \r\n" );
document.write( "3 to column 2, so the 'carry' to column 2 must be 1.  So we\r\n" );
document.write( "put a 1 above column 2:\r\n" );
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document.write( "1234\r\n" );
document.write( "↓↓↓↓\r\n" );
document.write( " 1 1\r\n" );
document.write( "  AA\r\n" );
document.write( "+BCB\r\n" );
document.write( "1EE1\r\n" );
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document.write( "The only digit that B could be is 9, because that is the only\r\n" );
document.write( "digit that a 'carry' of 1 to results in a 2-digit number.\r\n" );
document.write( "And that 2 digit number is, of course 10.  So we know that \r\n" );
document.write( "B=9 and E=0, so we replace those letters by digits:\r\n" );
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document.write( "1234\r\n" );
document.write( "↓↓↓↓\r\n" );
document.write( " 1 1\r\n" );
document.write( "  AA\r\n" );
document.write( "+9C9\r\n" );
document.write( "1001\r\n" );
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document.write( "Column 4 tells us that A=2. So we have\r\n" );
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document.write( "1234\r\n" );
document.write( "↓↓↓↓\r\n" );
document.write( " 1 1\r\n" );
document.write( "  22\r\n" );
document.write( "+9C9\r\n" );
document.write( "1001      \r\n" );
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document.write( "Now C has to cause column 3 to add up to 10,\r\n" );
document.write( "So C=7, and the addition is\r\n" );
document.write( " 1 1\r\n" );
document.write( "  22\r\n" );
document.write( "+979\r\n" );
document.write( "1001\r\n" );
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document.write( "Edwin
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