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Problem 48

The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.

Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
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euler48 = (sum $ map (\x -> x^x) [1..1000]) `mod` 10^10


