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

  • math -

    let's look at the first few terms
    2(1!) = 2
    2(2!) = 4
    2(3!) = 12
    2(4!) = 48
    2(5!) = 240
    2(6!) = 1440
    2(7!) = 10080
    2(8!) = 80640
    2(9!) = 725760
    2(10!) = 7257600
    2(11!) = .....3600 , since we are only concerned about the last 3 digits
    2(12!) = ..... 3200
    2(13!) = ... 1600
    2(14!) = ...2400
    2(15!) = ...36000
    after that no result will contribute to the last 3 digits.

    so add them up , to the end of the first 3 digits.
    Unless I made a silly arithmetic error I got 626

  • math -

    N=2^(1!)!+2^(2!)!+...+2^(1000!)!?
    this is the actual series..

  • math -

    stop this.

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