(數學)高一數學問題...

1. 22007除以10的餘數?
[解]
(mod 10)
2^1≡2，2^2≡4，2^3≡8，2^4≡16≡6，
2^5≡(2^4)*2≡(6)*2≡12≡2，2^6≡(2^5)*2≡(2)*2≡2^2，…
由上可看出2^n除以10的餘數，是四次一循環
而2007除以4餘3，所以2^2007≡2^3≡8


2. 若n為正整數,且n4-27n2+81為質數p ,則數對(n,p)=?
[解]
n^4 - 27n^2+81
= (n^4+81) - 27n^2
= (n^2-9)^2 - 27n^2 + 18n^2
= (n^2-9)^2 - 9n^2
= (n^2-9)^2 - (3n)^2
= [(n^2-9)-3n]*[(n^2-9)+3n]

因為是質數p，所以n^2-9-3n=1，n^2-9+3n=p
=> n^2 - 3n - 10 = 0
=> (n + 2)(n - 5)=0
=> n=-2(不合) 或 n=5
所以p = n^2 - 9 + 3n = 25 - 9 + 15 = 31
∴(n,p) = (5, 31)


3. 正整數序對(x,y)滿足xy+6x-11=x2 ,則(x,y)=?
[解]
=> xy+6x-x^2 = 11
=> x(y+6-x) = 11
=> x=1 或 x= 11

(1)x=1時：y+6-x=11 => y+6-1=11 => y=6 => (x,y)=(6,6)
(2)x=11時：y+6-x=1 => y+6-11=1 => y=6 => (x,y)=(11,6)
∴Ans：(x,y) = (6,6) , (11,6)


2009-08-20 18:38:13 補充：
這裡有同餘的一些定義與性質介紹：
http://tw.knowledge.yahoo.com/question/question?qid=1008102102445

2009-08-20 21:29:44 補充：
更正：
(1)x=1時：y+6-x=11 => y+6-1=11 => y=6 => (x,y)=("1",6)
(2)x=11時：y+6-x=1 => y+6-11=1 => y=6 => (x,y)=(11,6)
∴Ans：(x,y) = ("1",6) , (11,6)

竟然被說粗心了，orz

1 2^2007 (mod 10)

=(2^400)(2^7) (mod 10)

=(2^5)^80(8) (mod 10)

=8

So the remainder is 8

2 n4-27n2+81

=(n^2-9)^2-9n^2

=(n^2-9)^2-(3n)^2

=(n^2+3n-9)(n^2-3n-9)

Since this is a prime number, either (n^2+3n-9)=1 or (n^2-3n-9)=1

=>n=5,p=31

3 xy+6x-11=x^2

y+6-11/x=x

So 11/x=>x=1 or 11

(x,y)=(1,6);(11,6)

這是作業題嗎? 我怎麼哪邊看過

mod 為餘數的意思
另外，轉貼一下樓上的解答，然後更正一下，不好意思，樓上的有點粗心喔

1 2^2007 (mod 10)

=(2^400)(2^7) (mod 10)

=(2^5)^80(8) (mod 10)

=(2^80)(8) (mod 10)

=(2^16)(8) (mod 10) =[(2^5)^3 x 2 x 8] mod 10

=[(2^3) x 2 x 8 ] mod 10

=128 mod 10 = 8

So the remainder is 8


2 n4-27n2+81

=(n^2-9)^2-9n^2

=(n^2-9)^2-(3n)^2

=(n^2+3n-9)(n^2-3n-9)

Since this is a prime number, either (n^2+3n-9)=1 or (n^2-3n-9)=1

因為 n^2+3n-9 > n^2-3n-9，所以 n^2-3n-9=1
=>n=5 or -2(不成立)
=>p=31

A: (n,p)=(5,31)

第三題正確
3 xy+6x-11=x^2

y+6-11/x=x

So 11/x=>x=1 or 11

when x=1,y=6

when x=11,y=(121+11-66)/11=11+1-6=6

(x,y)=(1,6);(11,6)