I don't feel like this is a easy question


@tankztc said in I don't feel like this is a easy question:
This question just doesn't sound like an easy one to me... ╮(╯﹏╰)╭
LOL, neither do I.



Seems that if n is an even number, you can guess out the largest palindromic number.
n=2:
101*99=99,99; take out the middle 99, which is 990/99=10, then it is (10110)*99=91*99=9009n=4:
10001*9999=9999,9999, take out the middle 9999, which is 999900/9999=100, then it is (10001100)*9999=9901*9999=99000099and so on.

@tankztc The reason why this is easy is that there is a straight forward solution, even though that might have the worst case complexity. O(10^n)^2.