SOLUTION: 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 whe

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?
Found 2 solutions by richwmiller, Edwin McCravy:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
the third must end in 1
1xx1
first must be one of these
11,22,33,44,55,66,77,88,99
second must be
1y1,2y2,3y3
4y4,5y5,6y6
7y7,8y8,9y9
the first and second must add up a number ending in 1
the first and second must add up a number beginning with 1
the first and second must add up to a number greater than 1000
9 and 2
greater 901 plus 99 is a four digit number
since it begins with 1 it must end in 1
5y5+66 is not 4 digits
so it must 9y9 and 22=1xx1
909+22+10x=1001+z
931+10a=1001+b,
b=100c+10c
c<10
a<10
b>10
c=0
999+22=1021
989+22=1011
979+22=1001
bingo

Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!
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?

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 ↓↓↓↓ AA +BCB 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 ↓↓↓↓ AA +BCB 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 ↓↓↓↓ ₁ AA +BCB 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 ↓↓↓↓ _{1 1} AA +BCB 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 ↓↓↓↓ _{1 1} AA +9C9 1001 Column 4 tells us that A=2. So we have 1234 ↓↓↓↓ _{1 1} 22 +9C9 1001 Now C has to cause column 3 to add up to 10, So C=7, and the addition is _{1 1} 22 +979 1001 Edwin

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