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Palindromes


Date: 02/24/98 at 13:46:35
From: Anonymous
Subject: Palindromes

Two 4-digit palindromes are added to give a 5-digit palindrome N. 
What are the possible numbers for N?


Date: 02/24/98 at 17:34:57
From: Doctor Sam
Subject: Re: Palindromes

Keshni,

Suppose we represent the digits of the palindrome with letters. Your 
problem becomes:

       a  b  b  a 
    +  c  d  d  c
    -------------
    e  f  g  f  e

The largest possible four-digit number is 9999, which is less than 
10,000.  So abba + cddc must be less than 20,000.  So e = 1.

Now we know that e = 1, so the units digit of the sum is one. How can 
a+c give 1?  Neither can be zero (since we cannot start a number with 
0) so it must be that a+c = 11.  

Now if a+c = 11 in the units place, then a+c also gives 11 in the 
thousands place. What does that tell us about f?

Either there is a carry from the b+d addition in the hundreds place or 
not. This gives two possibilities:

NO CARRY from b+d                  CARRY from b+d

       a  b  b  a                      a  b  b  a
    +  c  d  d  c                   +  c  d  d  c
    -------------                   -------------
    1  1  g  f  1                   1  2  g  f  1

This gives us a value for f:

(1)    a  b  b  a                 (2)  a  b  b  a
     + c  d  d  c                    + c  d  d  c
     ------------                   -------------
     1 1  g  1  1                   1  2  g  2  1

Now what?  In case (1) there is no carry from b+d and a+c = 1.  
In this case the addition in the tens place is b+d+1 = 1 or 11.   
The 11 answer is not a possibility here because that would make 
b+d = 10 and we are assuming that there is no carry from b+d.  
So the only possibility is b+d = 0, which only happens if b = 0 
and d = 0.  In this case g = 0 as well and there are four 
possibilities:

        2 0 0 2        3 0 0 3         4 0 0 4          5 0 0 5
      + 9 0 0 9      + 8 0 0 8       + 7 0 0 7        + 6 0 0 6
      ---------      ---------       ---------        ---------
      1 1 0 1 1      1 1 0 1 1       1 1 0 1 1        1 1 0 1 1  

So in this case N = 11011

That leaves case (2) where  b+d produces a carry.
 
CARRY from b+d

        a  b  b  a
     +  c  d  d  c
     -------------
     1  2  g  f  1

Since this is a palindrome, f = 2. Since a+c = 11, we can now look at 
the addition in the tens place:  b + d + 1 = 2 or 12  (since f = 2).

case (2a):  b+d+1=2 so b+d = 1.  This can only happen if b = 0 and 
d = 1 or vice versa. This gives two more options:

        a  0  0  a              a  1  1  a
     +  c  1  1  c           +  c  0  0  c
     -------------           ------------- 
     1  2  1  2  1           1  2  1  2  1    

So N = 12121

Finally we have case (2b): b+d+1 = 12 or b+d = 11. In this case g = 2 
and so N = 12221.

Hope that helps.

-Doctor Sam,  The Math Forum
 Check out our web site http://mathforum.org/dr.math/   
    
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