[Toán 8] Bài khó

K

kool_boy_98

tìm số dư khi chia 1992^1993+1994^1995 cho 7

Ta thấy 199571995 \vdots 7, do đó:

19921993+19941995=(BS73)1993+(BS71)1995=BS731993+BS711992^{1993}+1994^{1995}=(BS 7-3)^{1993}+(BS 7-1)^{1995}=BS 7-3^{1993}+BS 7-1

Mặt khác ta có: 31993=36k+1=3.(33)2k=3.(BS71)2k=3.(BS7+1)=BS7+13^{1993}=3^{6k+1}=3.(3^3)^{2k}=3.(BS 7-1)^{2k}=3.(BS 7 + 1) = BS 7 + 1, nên:

19921993+19941995=BS7(BS73)1=BS741992^{1993}+1994^{1995}=BS 7 -(BS 7 - 3) - 1=BS 7 -4 chia cho 7 dư 3

Vậy ....
 
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