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www.mikescher.com/www/statics/euler/Euler_Problem-063_explanation.md

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2017-11-08 17:39:50 +01:00
After all the previous programs, this one is surprisingly ... dense (the main code-block is 54x8).
The algorithm is quickly explained for each length n we calculate the numbers `1^n`, `2^n` ... until `9^n` and see which have a length of `n`.
(From `10^n` upwards the condition is impossible, because `10^n` has `(n+1)` digits).
The main problem is that the numbers exceed Int64. So we need to implement long multiplication ... again. (see problem 16, 20, 29, 56 and 57)