Created
August 27, 2015 10:32
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| # Find the whole number _n_ that consists of 10 digits | |
| # (e.g. 3451209876) for which the first _k_ digits | |
| # can be divided by _k_ with `2 <= _k_ <= 9` | |
| # | |
| # Examples: | |
| # The first 2 digits of _n_ can be divided by 2 (`34` => true) | |
| # The first 3 digits of _n_ can be divided by 3 (`345` => true) | |
| # The first 4 digits of _n_ can be divided by 4 (`3451` => false) | |
| # etc. | |
| # | |
| # How many can you find? | |
| def magic_number?(candidate) | |
| return false if candidate.size < 10 | |
| return false if candidate.join.to_i % 10 != 0 | |
| return false if candidate[0,9].join.to_i % 9 != 0 | |
| return false if candidate[0,8].join.to_i % 8 != 0 | |
| return false if candidate[0,7].join.to_i % 7 != 0 | |
| return false if candidate[0,6].join.to_i % 6 != 0 | |
| return false if candidate[0,5].join.to_i % 5 != 0 | |
| return false if candidate[0,4].join.to_i % 4 != 0 | |
| return false if candidate[0,3].join.to_i % 3 != 0 | |
| return false if candidate[0,2].join.to_i % 2 != 0 | |
| true | |
| end | |
| (0..9).to_a.permutation.each do |candidate| | |
| puts candidate.join.to_i if magic_number? candidate | |
| end |
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This was written for a math assignment. I wonder if it can be made more efficient (both in terms of byte size and speed).