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www.mikescher.com/www/statics/euler/Euler_Problem-023_explanation.md

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2017-11-08 17:39:50 +01:00
The algorithm here is pretty ineffecient. First I brute force all the abundent numbers and then calculate every possible product of these numbers.
I used a little trick to store 2 boolean values per grid cell: The lowest bit represents if the value is abundent and the second-lowest if it can be written as two abundent numbers