7 lines
557 B
Markdown
7 lines
557 B
Markdown
Here we iterate through the values of a (1..99) and do (manually) the long multiplication.
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Because we have to implement the multiplication manually we get every result from a^1 to a^99 an can easily get the digitsum for these.
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Then we remember the maximum digitsum and vòila, problem solved.
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A few optimizations:
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- We ignore values where `a%10 == 0`, because these result in numbers consisting mostly of zeroes, and those will never have a high digitsum
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- We also ignore values for a<45, because the numbers are just to short to be really significant. |