Re: [中學] 餘數
※ 引述《yk1224 (這是我們的紀念日)》之銘言：
: 2009
: 1.求3 除以1000之餘數為
By Euler's thm
3^(400)=1 mod 1000.
Therefore 3^2009=3^9=683 mod 1000
Rmk, if you know chinese remainder thm, you also can use it to solve it.
Let
x=3 mod 8
x=58 mod 125
Hint: 3^5=243=(-7) mod 125. Cosider (-7)^4 then you can get this.
Therefore, by Chinese remainder thm, you also can get the same conclusion.
: ANS:683
: 2000 4 3 2 3 2
: 2.若(x -1)除以(x + x + 2x + x + 1)之餘式為 ax + bx + cx +d，
: 則a+b+c+d之值?
: ANS: -6
: 第一題覺得應該是要從找規律下手
: 可是卻找不出來 囧 麻煩高手提點一下要怎麼處理
: 第二題則看到就卡了 哭哭
: 先感謝各位解惑囉!
Recall x^2+1 |x^12-1 and x^2+x+1|x^12-1 =>x^4+x^3+2x^2+x+1|x^12-1
It implies that x^2000-1=x^8-1 mod x^12-1
Therefore, x^8-1=(x^4+x^3+2x^2+x+1)(x^3-x-x+2)+-2x^3-x^2-2x-1.
So a+b+c+d=-6
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