Violympic toán 8

H24

Tìm dư của : a, 31993 khi chia cho 7

b, 19921993+19941995 khi chia cho 7

ST
17 tháng 8 2018 lúc 11:07

a) Dư 3

b) Dư 7

Bình luận (2)
TH
17 tháng 8 2018 lúc 20:38

a)

\(3^{1993}:7\)

Ta có:

\(3\equiv3mod\left(7\right)\)

\(3^3\equiv6mod\left(7\right)\)

\(3^9\equiv6^3\equiv6mod\left(7\right)\)

\(3^{10}\equiv6.3\equiv4mod\left(7\right)\)

\(3^{90}\equiv1^3\equiv1mod\left(7\right)\)

\(3^{100}\equiv1.4\equiv4mod\left(7\right)\)

\(3^{900}\equiv4^9\equiv1mod\left(7\right)\)

\(3^{1000}\equiv1.4\equiv4mod\left(7\right)\)

\(3^{1993}\equiv4.1.1.6\equiv3mod\left(7\right)\)

Vậy số dư của \(3^{1993}\) khi chia cho \(7\) là 3

b)

\(1992^{1993}+1994^{1995}\) \(:7\)

Ta có:

\(1992\equiv4mod\left(7\right)\)

\(1992^3\equiv4^3\equiv1mod\left(7\right)\)

\(1992^9\equiv1^3\equiv1mod\left(7\right)\)

\(1992^{10}\equiv1.4\equiv4mod\left(7\right)\)

\(1992^{90}\equiv4^9\equiv1mod\left(7\right)\)

\(1992^{100}\equiv1.4\equiv4mod\left(7\right)\)

\(1992^{900}\equiv4^9\equiv1\)\(mod\left(7\right)\)

\(1992^{1000}\equiv1.4\equiv4mod\left(7\right)\)

\(1992^{1993}\equiv4.1.1.1\equiv4mod\left(7\right)\)

Ta có:

\(1994\equiv6mod\left(7\right)\)

\(1994^5\equiv6^5\equiv6mod\left(7\right)\)

\(1994^{10}\equiv6^2\equiv1mod\left(7\right)\)

\(1994^{90}\equiv1^9\equiv1mod\left(7\right)\)

\(1994^{100}\equiv1.1\equiv1mod\left(7\right)\)

\(1994^{900}\equiv1^9\equiv1mod\left(7\right)\)

\(1994^{1000}\equiv1.1\equiv1mod\left(7\right)\)

\(1994^{1995}\equiv1.1.1.6\equiv6mod\left(7\right)\)

Ta có:

\(1992^{1993}+1994^{1995}\)\(=4+6\equiv3mod\left(7\right)\)

Vậy \(1992^{1993}+1994^{1995}\)chia cho 7 có số dư là 3

Chúc bạn học tốt!

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