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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 6: tìm số dư khi chia 2005^2021 cho 21

Toán Lớp 6: tìm số dư khi chia
2005^2021 cho 21

Comments ( 1 )

  1. Giải đáp: $2005^{2021}$ chia $21$ dư $14$
    Lời giải và giải thích chi tiết:
    Ta có:
    $2005\equiv 1(mod 3)$
    $\to 2005^{2021}\equiv 1^{2021}(mod 3)$
    $\to 2005^{2021}\equiv 1(mod 3)$
    $\to 2005^{2021}=3k+1, k\in N$
    Lại có:
    $2005\equiv 3(mod 7)$
    $\to 2005^{2021}\equiv 3^{2021}(mod 7)$
    Mà $3^2\equiv 2(mod 7)$
    $\to (3^2)^3\equiv 2^3(mod 7)$
    $\to 3^6\equiv 8(mod 7)$
    $\to 3^6\equiv 1(mod 7)$
    $\to (3^6)^{336}\equiv 1^{336}(mod 7)$
    $\to 3^{2016}\equiv 1(mod 7)$
    $\to 3^{2016}\cdot 3^5\equiv 3^5(mod 7)$
    $\to 3^{2021}\equiv 3^5\equiv 5(mod 7)$
    $\to 2005^{2021}\equiv 5(mod 7)$
    $\to 2005^{2021}=7q+5, q\in N$
    $\to 3k+1=7q+5$
    $\to 3k=7q+4$
    $\to q$ chia $3$ dư $2$
    $\to q=3p+2, p\in N$
    $\to 2005^{2021}=7q+5=7(3p+2)=21p+14$
    $\to 2005^{2021}$ chia $21$ dư $14$

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222-9+11+12:2*14+14 = ? ( )

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