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Đề:Tìm số dư của:
a)$2^{2150}$ cho $5$;
b)$3^{1993}$cho $7$
Giải:
a)$2^{10}=1024 \equiv -1 \pmod{5}$@};-
\Leftrightarrow $ (2^{10})^5=2^{50} \equiv (-1)^5 \pmod{5} \equiv -1 \pmod{5}$
\Leftrightarrow $(2^{50})^{43} \equiv (-1)^{43} \pmod{5} \equiv -1 \pmod{5} \equiv 4 \pmod{5} $


\Leftrightarrow $2^{2150} \equiv 4 \pmod{5}$
Vậy $2^{2150}$ chia cho 5 dư 4
b)$3^{1993}=3^{1992}.3$

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*$3^3 \equiv -1 \pmod{7}$


\Leftrightarrow $(3^3)^664 \equiv(-1)^664 \pmod{7} \equiv 1 \pmod{7}$
\Leftrightarrow $3^{1992} \equiv 1 \pmod{7}$
\Rightarrow $3^{1992}$ chia 7 dư 1
*$3$ chia 1 dư 3/

\Rightarrow$3^{1993}=3^{1992}.3$ chia cho 7 dư $1.3=3$



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