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find( without using calculators or computer programs) the last two decimal digits of : (2007)^2007?

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  • Anonymous
    1 decade ago
    Favorite Answer

    To find the last two digits of 2007^2007,

    we will have to determine the pattern with the last digits of 2007^n......

    2007^1 = 2007

    Last 2 Digits Of 2007^1 --> 07

    2007^2 = 4 028 049

    Last 2 Digits Of 2007^2 --> 49

    2007^3 = 8 084 294 343

    Last 2 Digits Of 2007^3 --> 43

    2007^4 = 16 225 178 746 401

    Last 2 Digits Of 2007^4 --> 01

    2007^5 = 32 563 933 744 026 807

    Last 2 Digits Of 2007^5 --> 07

    2007^6 = 65 355 815 024 261 801 649

    Last 2 Digits Of 2007^6 --> 49

    2007^7 = 131 169 120 753 693 435 909 543

    Last 2 Digits Of 2007^7 --> 43

    2007^8 = 263 256 425 352 662 725 870 452 801

    Last 2 Digits Of 2007^8 --> 01

    By observation,

    when n is a multiple of 4, the last digits are 01.

    when (n + 1) is a multiple of 4, the last digits are 43.

    when (n + 2) is a multiple of 4, the last digits are 49.

    when (n + 3) is a multiple of 4, the last digits are 07.

    Therefore, the last 2 digits of 2007^2007 is 43 since

    (2007 + 1) is divisible by 4.

  • 1 decade ago

    Use cyclicity man its repeats at

    7^1 units = 07

    7^2 units = 49

    7^3 units = 43

    7^4 = 01

    7^5 = 07

    7^6 = 49

    7^7 = 43

    cyclicity =04

    and since the 2007 ^2007 has zeroes the units and tens places shall be 43(343)

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