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發表 aaddfg 於 星期五 十一月 03, 2006 8:56 pm

#ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=name#ed_cl##ed_op#A name=""#ed_cl##ed_op#/A#ed_cl##ed_op#FONT size=2#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#SPAN class=name#ed_cl##ed_op#FONT size=2#ed_cl#補充一下#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=name#ed_cl##ed_op#FONT size=2#ed_cl#滿足(x,12)=1意思就是x不能有2,3的因數#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=name#ed_cl##ed_op#FONT size=2#ed_cl#所有的數-2的倍數-3的倍數+6的倍數#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=name#ed_cl##ed_op#FONT size=2#ed_cl#1+2+3+....+300-(2+4+6+.....+300)-(3+6+9+......+300)+(6+12+18+......+300)#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=name#ed_cl##ed_op#FONT size=2#ed_cl#=300*301/2-150*302/2-100*303/2+50*306/2#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=name#ed_cl##ed_op#FONT size=2#ed_cl#=15000#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl#

發表 jackhoung 於 星期五 十一月 03, 2006 8:12 pm

#ed_op#DIV#ed_cl#(1+300)*300/2-(2+300)*150/2-(3+300)*100/2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#+(6+300)*50/2=15000#ed_op#/DIV#ed_cl#

[數學]等差級數

發表 skywalker 於 星期五 十一月 03, 2006 7:18 pm

#ed_op#DIV#ed_cl#若x∈N且1≤x≤300,則滿足(x,12)=1之一切x的總和#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#答案是7344,但我算的答案就是不一樣,想確認一下#ed_op#/DIV#ed_cl#