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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