Cody

# Reggie Wilcox

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 on 21 Oct 2017 on 21 Oct 2017 on 21 Oct 2017 on 20 Oct 2017 on 20 Oct 2017 on 20 Oct 2017 on 19 Oct 2017 My combinatoric approach and brute force method both agree for x = 10 with 1454400 and for x = 11 with 9072000. I can't brute force beyond that, but an reasonably I am handling the redundant permutations introduced when we have duplicate digits due to these two cases. For reference, I get 3216477600 for x = 14. on 19 Oct 2017 on 19 Oct 2017 on 19 Oct 2017 on 19 Oct 2017 I don't think the case for x > 9 is handled correctly. Consider x = 11. My interpretation is that a pandigital number of order 11 will contain exactly two 0's, two 1's and one of every digit 2 through 9. Using a brute force approach, I can check every pandigital number of order 11 and see if it is a multiple of 11: count = 0; digs = 0:11; for d0 = 1:9 strs = unique(perms(char(mod(setdiff(digs, d0), 10) + '0')), 'rows'); nums = str2num([repmat(char(d0 + '0'), size(strs, 1), 1) strs]); count = count + sum(mod(nums, 11) == 0); end Quite simply, this loops over each starting digit, except zero, takes all possible unique permutations of the remaining digits, then counts the number divisible by 11. This yields a solution of 9072000. Note that this approach is not a suitable answer for Cody, since its runtime is on the order of an hour, but is the simplest way to demonstrate a correct solution for x = 11. on 19 Oct 2017 on 19 Oct 2017 on 19 Oct 2017 on 18 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 17 Oct 2017 on 16 Oct 2017 on 16 Oct 2017 on 16 Oct 2017 on 16 Oct 2017 on 16 Oct 2017 on 16 Oct 2017 on 16 Oct 2017 on 16 Oct 2017 on 16 Oct 2017 on 16 Oct 2017 on 16 Oct 2017 on 16 Oct 2017 on 16 Oct 2017
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