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What are the last three digits of
N=(2^(1!)!) + (2^(2!)!) + ... + (2^(1000!)!)?

please help..

  • math -

    N = 2^1 + 2^2! + 2^(1000!)!)
    The last number has no ending digits as small as 2 or 4

    006 would be the last three digits of the complete number.

  • math -

    Since any number n! where n>=5 ends in zero,

    (2^1)! + (2^2)! + ...
    = 2 + 24 + ...
    ends in 6

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