數學題(TRML個人賽)兩題

回答 (4)

2014-03-25 5:48 pm
✔ 最佳答案
1設f(x)=(ax+b)(x-1)(x-2)(x-3)(x-4)+x+2
f(0)=b*24+2=1
b=-1/24
f(5)=(5a-1/24)*24+7=18
120a-1+7=18
120a=12
a=1/10
f(x)=(x/10-1/24)*(x-1)(x-2)(x-3)(x-4)+x+2
f(6)=(6/10-1/24)*5*4*3*2+8=75

2
(n^2+7)/(2n+3)為整數
(4n^2+28)/(2n+3) 為整數
(4n^2+28)/(2n+3)=2n-3+37/(2n+3)
(2n+3)|37
最大2n+3=37
n=17
(17^2+7)/(2*17+3)=8 (ok)


2014-05-30 5:32 pm
參考下面的網址看看

http://phi008780520.pixnet.net/blog
2014-03-25 3:49 pm
2n+3整除n^2+7
2n+3整除4n^2+28
2n+3整除2n(2n+3)-6n+28
2n+3整除2n(2n+3)-3(2n+3)+9+28
2n+3整除2n(2n+3)-3(2n+3)+37
2n+3整除37
若2n+3=37,n=17
驗算
2n+3=37,n^2+7=296,

前半段,可用多項式除法做
關鍵在於先乘4倍,讓它好做除法
2014-03-25 8:46 am
I-3 我算答案是 75.


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