YLL討論網

#ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#真是太久沒來這個網站 (我們家的電腦有點......)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#現在給第三題的答案......#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#令費氏數列中的第 n 項為 F#ed_op#SUB#ed_cl#n#ed_op#/SUB#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以連八項的和為 F#ed_op#SUB#ed_cl#k #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+1 #ed_op#/SUB#ed_cl#+ ......#ed_op#SUB#ed_cl# #ed_op#/SUB#ed_cl#+ #ed_op#STRONG#ed_cl##ed_op#EM#ed_cl#F#ed_op#SUB#ed_cl#k+6 #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+7#ed_op#/SUB#ed_cl##ed_op#/EM#ed_cl##ed_op#/STRONG#ed_cl# > F#ed_op#SUB#ed_cl#k+8#ed_op#/SUB#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#而 (F#ed_op#SUB#ed_cl#k #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+1#ed_op#/SUB#ed_cl#)#ed_op#SUB#ed_cl# #ed_op#/SUB#ed_cl#+ (F#ed_op#SUB#ed_cl#k+2 #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+3#ed_op#/SUB#ed_cl#)#ed_op#SUB#ed_cl# #ed_op#/SUB#ed_cl#+ (F#ed_op#SUB#ed_cl#k+4 #ed_op#/SUB#ed_cl#+.F#ed_op#SUB#ed_cl#k+5#ed_op#/SUB#ed_cl#)#ed_op#SUB#ed_cl#  #ed_op#/SUB#ed_cl#+ (F#ed_op#SUB#ed_cl#k+6 #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+7#ed_op#/SUB#ed_cl#) #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=   F#ed_op#SUB#ed_cl#k+2 #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+4 #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+6 #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+8#ed_op#/SUB#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#<   #ed_op#STRONG#ed_cl##ed_op#EM#ed_cl#F#ed_op#SUB#ed_cl#k+3 #ed_op#/SUB#ed_cl#+ F#ed_op#/EM#ed_cl##ed_op#/STRONG#ed_cl##ed_op#SUB#ed_cl##ed_op#STRONG#ed_cl##ed_op#EM#ed_cl#k+4#ed_op#/EM#ed_cl##ed_op#/STRONG#ed_cl# #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+6 #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+8#ed_op#/SUB#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=   #ed_op#STRONG#ed_cl##ed_op#EM#ed_cl#F#ed_op#SUB#ed_cl#k+5 #ed_op#/SUB#ed_cl#+ F#ed_op#/EM#ed_cl##ed_op#/STRONG#ed_cl##ed_op#SUB#ed_cl##ed_op#STRONG#ed_cl##ed_op#EM#ed_cl#k+6#ed_op#/EM#ed_cl##ed_op#/STRONG#ed_cl# #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+8#ed_op#/SUB#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=   #ed_op#STRONG#ed_cl##ed_op#EM#ed_cl#F#ed_op#SUB#ed_cl#k+7 #ed_op#/SUB#ed_cl#+ F#ed_op#SUB#ed_cl#k+8#ed_op#/SUB#ed_cl##ed_op#/EM#ed_cl##ed_op#/STRONG#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=   #ed_op#EM#ed_cl##ed_op#STRONG#ed_cl#F#ed_op#SUB#ed_cl#k+9#ed_op#/SUB#ed_cl##ed_op#/STRONG#ed_cl##ed_op#/EM#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#因為原式必大於F#ed_op#SUB#ed_cl#k+8#ed_op#/SUB#ed_cl#小於F#ed_op#SUB#ed_cl#k+9#ed_op#/SUB#ed_cl# ,得證#ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#可以請aa2191943繼續出題嗎??而且可否給出fibonacci那題的解法???#ed_op#/DIV#ed_cl#

這題我會
4.令a=,b=
原式改為
<1>若a+b不為0
同乘以得ab=1
即
化簡：x=0,99
<2>若a+b=0

化簡：
故
--------------------------------------------
共三組解

#ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#4. 試求下列方程之實數解:&nbsp;&nbsp;&nbsp; ( 難度: 十年級 )#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;&nbsp; ( x - 50 ) / 49 + ( x - 49 ) / 50 = 49 / ( x - 50 ) + 50 / ( x - 49 )#ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#95.08.12.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#3. proof : 費不納西數列中連續八項之和不可為數列中的其他項#ed_op#/DIV#ed_cl#

莫斯科數學奧林匹克考費馬數!?
那這應該是很高竿的比賽囉...
還有其他題目嗎? (摩拳擦掌ing~)

#ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#參考:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(1)若 n &gt; 1, 但不是 2 的乘冪, 則 n#ed_op#SUP#ed_cl#n #ed_op#/SUP#ed_cl#+1 必為合數#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(&nbsp; 因為a#ed_op#SUP#ed_cl#n#ed_op#/SUP#ed_cl#+1=(a+1)(a#ed_op#SUP#ed_cl#n-1#ed_op#/SUP#ed_cl#-a#ed_op#SUP#ed_cl#n-2#ed_op#/SUP#ed_cl#+a#ed_op#SUP#ed_cl#n-3#ed_op#/SUP#ed_cl#- ...... +a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-a+1)&nbsp; )#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(2)故 n 只能取1,2,4,8,16#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(3)明顯的,n=1,2,4合, 8不合#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(4) 16#ed_op#SUP#ed_cl#16 #ed_op#/SUP#ed_cl#+ 1 = 2#ed_op#SUP#ed_cl#64 #ed_op#/SUP#ed_cl#+ 1 = 2^(2#ed_op#SUP#ed_cl#6#ed_op#/SUP#ed_cl#)#ed_op#SUB#ed_cl# #ed_op#/SUB#ed_cl#+1, 已知費馬數, n = 6為合數#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以答案: n = 1,2,4#ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

這個我知道...
但是因為那樣子分解對此題沒有什麼幫助.所以才那樣說的.

不過這題又到底是要怎樣解才算完美呢?

tangpakchiu 寫到:#ed_op#DIV#ed_cl#宇智波鼬，可以告訴x^n+y^n，和x^n-y^n的公式和n有什麼限制給我知嗎???#ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#原則上次方為偶數的數字之和無法分解成純實數相乘，但的確存在其分解方式:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT size=4#ed_cl##ed_op#STRONG#ed_cl#EX:a&sup2;+b&sup2;=(a+b√-1)(a-b√-1)#ed_op#/STRONG#ed_cl##ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT size=2#ed_cl#很明顯的，這種分解方式不是很實用...#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#那aa2191943可以公佈solution嗎???#ed_op#/DIV#ed_cl#

剛剛上網找到了.....16^16+1不是質數...
2^64 + 1 = 18446744073709551617 = 274177 × 67280421310721

http://zh.wikipedia.org/wiki/费马质数

tangpakchiu 寫到:(這句話不存在)怎麼想到mod37???而且我還有些地方不明白:為什麼8^8=2^24=(2^8)^3，所以不合????而且我還想知道是否一個數不被37整除就代表它是質數呢???x^n+y^n，我記得應有公式...忘了~~~~

要把數字刪掉,就要盡量證明他可以被某數整除(同餘0)
先講一下-1^(2n-1)=-1(就是說-1的奇次方=-1)
而-1+1=0
故我取37,才可以讓6^6餘-1
8^8+1=2^24+1=(2^8)^3+1一定可以被2^8+1整除,用的就是上面的原理

至於說16^16次方,老實講我也不是很肯定,但是若以我這個方法的話是求不到他有其他的因數,但也不能證明他是質數(汗...

tangpakchiu 寫到:而且我還想知道是否一個數不被37整除就代表它是質數呢???

回:並不是的...8並不能被37整除他也不是質數

8^8+1=(2^8)^3+1=(2^8+1)(2^16-2^8+1) 所以不合.


(n可為任意數)


n=2^m次時不可分解. (抱歉之前說成不能是偶數)

#ed_op#DIV#ed_cl#宇智波鼬，可以告訴x^n+y^n，和x^n-y^n的公式和n有什麼限制給我知嗎???#ed_op#/DIV#ed_cl#

是呀...我也有點不解.
怎證16^16+1是質數?

x^n+y^n的公式是有的...但n為奇數.

#ed_op#P#ed_cl#

aaddfg 寫到:第一題#ed_op#BR#ed_cl#首先淘汰掉除了1以外的奇數(奇數的奇次方+1為偶數)#ed_op#BR#ed_cl#剩下1,2,4,6,8,10,12,14,16,18#ed_op#BR#ed_cl#看看6#ed_op#BR#ed_cl#6^2　　　≡-1　(37)#ed_op#BR#ed_cl#(6^2)^3 ≡-1　(37)#ed_op#BR#ed_cl#6^6+1　 ≡0 　(37)#ed_op#BR#ed_cl#所以6也去掉了#ed_op#BR#ed_cl#同理可刪掉6,10,12,14,18#ed_op#BR#ed_cl#剩下1,2,4,8,16#ed_op#BR#ed_cl#其中8^8=2^24=(2^8)^3故不合#ed_op#BR#ed_cl#故答案為1,2,4,16#ed_op#BR#ed_cl##ed_op#BR#ed_cl#我用同餘...如何用因式分解??

#ed_op#/P#ed_cl##ed_op#P#ed_cl#aaddfg很強ar.....怎麼想到mod37???而且我還有些地方不明白:為什麼8^8=2^24=(2^8)^3，所以不合????而且我還想知道是否一個數不被37整除就代表它是質數呢???x^n+y^n，我記得應有公式...忘了~~~~#ed_op#BR#ed_cl##ed_op#/P#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

第一題
首先淘汰掉除了1以外的奇數(奇數的奇次方+1為偶數)
剩下1,2,4,6,8,10,12,14,16,18
看看6
6^2　　　≡-1　(37)
(6^2)^3  ≡-1　(37)
6^6+1　  ≡0 　(37)
所以6也去掉了
同理可刪掉6,10,12,14,18
剩下1,2,4,8,16
其中8^8=2^24=(2^8)^3故不合
故答案為1,2,4,16

我用同餘...如何用因式分解??

#ed_op#DIV#ed_cl#嗯, 第二題完全正確......#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#第一題可應用因式分解......#ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#第二題#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#可看出原式為 (xz-1)(xt-1)=x^3+1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#當x=1時，(x,y,z,t)=(1,5,2,3),(1,5,3,2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#　x=2時，(x,y,z,t)=(2,2,2,2),(2,3,1,5),(2,3,5,1)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#　x=3時，(x,y,z,t)=(3,2,1,5),(3,2,5,1)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#　x=5時，(x,y,z,t)=(5,1,2,3),(5,1,3,2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#&nbsp;#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#找不到其他解了#ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#其實這兩題都不會太難, 是中間程度的題目,#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#大家都可以做做看, 做不出來我再公佈答案#ed_op#/DIV#ed_cl#