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Let us consider a number, e.g. 1234

Now reverse the positions of the terminal digits, so we get 4231

4231-1234=2997 which is divisible by 9

i have seen this for n-digit numbers, where n ranges from 2 to 6

Is there any number which is contradicting this behavior irrespective of its number of digits?

Now coming back to the example,

Now 2997/9=333 is a palindrome

Another example 923456781-123456789=799999992

799999992/9=88888888

Is this also independent of the number of digits of the original number?

Let us consider :

95623-35629=59994/9=6666

The largest prime factor of 6666 is 101 which is a palindrome and a prime number.

Now 6666/101=66

Again 66/11(11 being the largest prime factor of 66) equals to 6 which is a palindrome.

Another example :

923456781-123456789=799999992/9=88888888

The largest prime factor of 88888888 is 137

88888888/137=648824/101(101 being largest prime factor of 648824)=6424/73(73 being largest prime factor of 6424) this in turn equals to 88 (a palindrome)

Finally, 88/11(11 being largest prime factor of 88 and a palindrome) equals to 8 which is also a palindrome.

Now as per (http://math.stackexchange.com/questions/200835/are-there-infinitely-many-super-palindromes), will it be correct to call 88888888 a super palindrome?

Can it be possible to call a number a super palindrome if it generates non palindromes along with palindromes, provided the non-palindromes finally generate using the method of division by largest prime factor, a palindrome other than 1?

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