#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#n+2??抱歉不大懂你的意思#ed_op#/DIV#ed_cl# #ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl# #ed_op#/FONT#ed_cl##ed_op#DIV#ed_cl#n = 95#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#因為 n+2 要和所有可能的 k(從19到91)互質??#ed_op#/DIV#ed_cl##ed_op#/SPAN#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#
#ed_op#br#ed_cl#不明....照道理最後一個數好像應該是91/(n+73) 或 111/(n+93)#ed_op#br#ed_cl#skywalker 寫到:#ed_op#div#ed_cl#求出最小的正整數n,使得下列分數每一個都是最簡分數#ed_op#/div#ed_cl##ed_op#div#ed_cl# #ed_op#/div#ed_cl##ed_op#div#ed_cl#19/(n+1),20/(n+2),.....,91/(n+93)#ed_op#/div#ed_cl#