Given a number N containing even number of digits. The task is to check whether that number is palindrome or not.
Input: N = 123321 Output: Palindrome Input: 1234 Output: Not palindrome
A Naive Approach is to traverse from front and back of that number and stop where they do not match.
An Efficient Approach is to use the below fact:
Palindrome Number having even number of digits is always divisible by 11.
Suppose the number is d1 d2 d3 d4…dn, whered1, d2, d3.. are digits of a number. If it is a palindrome then d1 = dn, d2 = dn-1, d3 = dn-2…..and so on. Now since divisibility of 11 states that the difference of sum of alternate digits of a number should be zero and same in the case of palindrome having even no. of digits i.e.
d1 + d3 + …+ dn-1 = d2 + d4 + d6 + … + dn
So, a palindromic number having even number of digits is always divisible by 11.
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