Palindrome Madness: Goodbye 2011, Hello 2012!

Maybe it’s because I’m in the midst of writing hundreds of problems for SUMiT 2012 that I feel like I’m seeing palindromic numbers everywhere! (Even in the date: 12/21…)

For those of us born in or before 1991, we’ve lived through two palindromic years: 1991 and 2002, and there won’t be another palindromic year for another century.

But did you ever happen to notice that if you take 2011, which isn’t a palindrome, and multiply it by its reversal, 1102, you get a palindrome? 2011 \times 1102 = 2216122.

That didn’t happen in 2010: 2010 \times 0102 = 205020, nor the year before that: 2009 \times 9002 = 18085018.

But…it happens again next year: 2012 \times 2102 = 4229224.

So, we’re about to live through two consecutive non-palindromic years whose product with their reversals is palindromic…something that won’t happen again until…any guesses?

It won’t happen again until the years 10011 and 10012! That’s a long time to wait.

This palindrome madness is all a bit silly anyway…after all, if you switch to, say, base 16, the palindromes go poof! (And if you play this game with the coefficients of a polynomial, the product will always be palindromic.) So let’s not attach undue significance to these pieces of mathematical trivia…it’s all in the lightness of the holiday spirit.

p.s. If you’d like a math problem to think about, here’s one: Suppose you pick a random whole number between 1 and 1,000,000 (inclusive) with all choices equally likely. What is the probability that the number is a palindrome? More generally, if the whole number is selected from between 1 and N, roughly how does the probability depend on N?

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