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Counting palindromes

December 19, 2009 pm31 9:56 pm

Dave at Math Notations put up a mildly interesting question about counting palindromes (length and another property defined). I just put up a comment, suggesting extensions…

Take a look. Comments on the original question?  But of more interest to me: do you have other ideas for extensions?

(question was: how many 5 digit palindromes with middle digit 0 are multiples of 9? As is his wont, Dave spoiled it by giving the answer up front. I approve. 10. He does this to focus attention on the discussion.

So far I offered two modifications:  1) get rid of the requirement that 0 be in the middle. Makes it a very different sort of counting problem, and far harder, imo. or 2) change “five digit” to less than 100,000. This raises a flurry of understanding questions, and forces 2 – 4 interesting clarifications to take place.

Go, look, leave your own comments and ideas.

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One Comment leave one →
  1. December 20, 2009 pm31 1:04 pm 1:04 pm

    I like both your extensions, and interestingly they were the same two that immediately appealed to me :).

    I think if I were to give this as a class-puzzle, it would probably run something like:

    A “palindromic number” is defined to be a (positive) integer which is unchanged upon reversal of its digits, such as 17871 or 5335. We do not allow leading zeroes, so 090 does not count.

    (A) Find the ten palindromic numbers with 5 digits and a zero in the centre (that is, of the form ab0ba) which are divisible by 9.

    (B) Find the ten palindromic numbers of the form a0b0a which are divisible by 9.

    (C) How many five-digit palindromic multiples of nine are there?

    (D) How many palindromic multiples of nine under 100 000 are there?

    (E) Under 1 000 000?

    This gives the weaker or less confident kids some attackable problems to get their teeth into before moving forward. For the stronger kids, (F) and (G) would be “can you PROVE your results?” and “how will you generalise” – just as always :).

    I might reverse A and B, though – I think B is actually slightly easier.

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