number theory

Find the remainder when 1!+2!+...+299!+300! is divided by 21

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  1. a) Find the last two digits(units and tens digit) in 1829^(1829)

    b) Find the units digit in 23^(7777)

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  2. To find large powers modulo some r, you can repeatedly square and/or multply numbers and reduce it modulo r at each step. E.g. if we do all coputations Mod 10 in the following, we can write:

    1829 = -1

    The 1829 th power of both sides is thus:

    1829^1829 = (-1)^1829 = -1

    Therefore the last digit of 1829^1829 is 9.

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  3. You can compute this Mod 100 using the Chinese Remainder Theorem, by ecaluating it separately Mod 25 and
    Mod 4. Mod 4 you have:

    1829 = 1, so

    1829^1829 = 1

    Mod 25 you have:

    1829 = 4

    Now 4^5 = 2^10 = 1024 = -1

    So, 4^10 = 1 and thus

    4^1829 = 4^9 = 4^(-1)

    4*6 = 24 = -1, so the inverse of 4 is -6 = 19

    We thus have that:

    1829^1829 = 1 Mod 4

    1829^1829 = -6 Mod 25

    We can then write down the answer Mod 100 as follows. If we denote the inverse of x mod y is denoted as
    [x^(-1)]_y, we can write down the solution as:

    1 * 25 * [25^(-1)]_4 +

    (-6) * 4 * [4^(-1)]_25

    Modulo 25 the first term is zero as it is multiple of 25, ad the second term is -6 as the 4 and the inverse of 4 mod 25 cancel. Modulo 4 the last term is zero as that isnow a multiple of 4 while the first term is 1 as 25 times the inverse of 25 mod 4 now cancel.

    We have:

    1 * 25 * [25^(-1)]_4 = 25

    (-6) * 4 * [4^(-1)]_25 = (-6)*4*(-6) = 24*6 = 44

    Note that in these computations we can reduce Mod 100.

    So, 1829^1829 Mod 100 = 25 + 44 = 69

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  4. 23^(7777) Mod 10 =

    3^7777 Mod 10

    Mod 10 we have:

    3^2 = 9 = -1 --->

    3^4 = 1

    7776 is a multiple of 4, so

    3^7777 = 3

    The last digit of 23^7777 is thus 3.

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