2 digit Palindrome describes a process by which most multi-digit numbers eventually become palindromes. The process being to repeatedly add a given number to the number formed when its digits are reversed.
For example: 152 -> 152+251=403 -> 403+304=707

I applied that process to a certain three digit number four times before I got a palindrome.
After the first and second additions, it was still a three digit number, and neither was a palindrome.
After the third addition it became a four digit number, but still not a palindrome.
And after the fourth addition the result was a four digit palindrome.

What number did I start with, assuming the first digit was smaller than the last?

I originally set this up in a spreadsheet and it quickly became obvious that the hundreds digit had to be "1" and that the units was "2".  It was then just a matter of deciding upon the tens value.

During review a note was made about there being more that one possibility. 
Unsure whether that reviewer had not immediately realised that there was a stipulation that the first digit was lower than the last.  However I wrote the following program allows for zero to be the first digit.

We would not normally consider a three-digit integer beginning with a leading zero to be three-digit.  That said the program finds 093 and 192 as solutions, both yielding 483 on the first addition.

It is howevr clear that the required three-digit is 192.



CLS
FOR a = 0 TO 8
FOR b = 0 TO 9
FOR c = a + 1 TO 9
n1 = a * 100 + b * 10 + c
n2 = c * 100 + b * 10 + a
n3 = n1 + n2
IF n3 < 1000 THEN
   d = INT(n3 / 100)
   e = INT((n3 - d * 100) / 10)
   f = n3 - d * 100 - e * 10
   IF d <> f THEN
     n4 = f * 100 + e * 10 + d
     n5 = n3 + n4
     IF n5 < 1000 THEN
       g = INT(n5 / 100)
       h = INT((n5 - g * 100) / 10)
       i = n5 - g * 100 - h * 10
       IF g <> i THEN
         n6 = i * 100 + h * 10 + g
         n7 = n5 + n6
         IF n7 > 1000 THEN
           j = INT(n7 / 1000)
           k = INT((n7 - j * 1000) / 100)
           l = INT((n7 - j * 1000 - k * 100) / 10)
           m = n7 - j * 1000 - k * 100 - l * 10
           IF j <> m AND k <> l THEN
             n8 = m * 1000 + l * 100 + k * 10 + j
             n9 = n7 + n8
             IF n9 < 10000 THEN
               n = INT(n9 / 1000)
               o = INT((n9 - n * 1000) / 100)
               p = INT((n9 - n * 1000 - o * 100) / 10)
               q = n9 - n * 1000 - o * 100 - p * 10
               IF n = q AND o = p THEN
                  PRINT a; b; c
                  PRINT n1; n2, n3; n4, n5; n6, n7; n8, n9
               END IF
             END IF
           END IF
         END IF
       END IF
     END IF
   END IF
END IF
NEXT
NEXT
NEXT

Posted by brianjn on 2009-11-27 22:17:05