Tìm số dư của phép chia

T

trinhminh18

ta có:$2004 \equiv 29$ (mod 1975)
\Rightarrow $2004^4 \equiv 29^4 \equiv 231$ (mod 1975)
$2004^6 \equiv 29^6 \equiv 721$ (mod 1975)
Tương tự như vậy:
$2004^{18} \equiv 1711$
$2004^{54} \equiv 1331$
$2004^{162} \equiv 1116$
$2004^{324} \equiv 1206$
$2004^{52}= (2004^{18})^2.(2004^6)^2.2004^4 \equiv 1711^2.721^2.231 \equiv 571.416.231 \equiv 1366$
\Rightarrow$2004^{376}=2004^{324}.2004^{52} \equiv 1206.1366 \equiv 246$ (mod 1975)
\Rightarrow$2004^{376}$ chia 1975 dư 246
Các bài sau làm tương tự
 
You must log in or register to reply here.
Top Bottom