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Post by Sentzeu » Thu Aug 26, 2004 6:18 pm

Hi to you all, this is my first post. Here's the first math puzzle that I've ever submitted.

Can any palindrom with an 2n number of digits be divided by 11?






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Post by Brown » Thu Aug 26, 2004 6:54 pm

Answer: Yes.

A test for divisibility by 11 is to add up the two sets of alternate digits in a number and see if they are identical or if they differ from each other by a factor of 11. If they do, the whole number is divisible by 11. For example, in determining whether 12,345,674 is divisible by 11, add 1+3+5+7 (the first, third, fifth and seventh digits) and compare to 2+4+6+4 (the second, fourth, sixth and eighth digits). In this case, both sums are identical, so 12,345,674 is divisible by 11.

In the case of a palindromic number having an even number of digits, the sums of alternate digits will always be the same.

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Post by Beleth » Thu Aug 26, 2004 7:01 pm


Bah, Brown beat me to it.

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Post by Sentzeu » Thu Aug 26, 2004 7:42 pm

I encountered this problem in a programming competetion.

We had to write a program that would find all palindromic primes up to 1000000. The first step we devised was to first generate a list of palindromes and then I suddenly realised that we only needed those with an odd number of digits. Counting off those who began with an even digit or 5 and we only needed to check 360 candidates.

Then after devising a quick prime checker we had ourselves a rather fast palindromic prime generator.

By the way the program had to be run on a 44Mhz processor platform within a certain timelimit and the language we made the program in was C, of course.