YLL討論網

糊塗迷蛋 寫到:#ed_op#DIV#ed_cl##ed_op#P#ed_cl#將千元鈔票的6位數字重新排列(由大到小)與原數字相減再檢去7所的數字不斷相加最後得2請問為什麼呀#ed_op#BR#ed_cl#EX 原數字:154111#ed_op#BR#ed_cl##ed_op#BR#ed_cl#541111#ed_op#BR#ed_cl#- 154111#ed_op#BR#ed_cl#=387000#ed_op#BR#ed_cl#- 7#ed_op#BR#ed_cl#=386993=38 =11 =2#ed_op#BR#ed_cl##ed_op#BR#ed_cl#3+8+6+9+9+3=38 3+8=11 1+1=2#ed_op#BR#ed_cl#最後降這6位數相加會等於2 #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#首先，這一題將減７的部分改成加２會更圓滿．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#現在來解這題：#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#HR#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#第一部分：兩個六位數相減．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#HR#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#在第一個相減的部分，我們知道相減的結果一定是９的倍數．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#為什麼？我們事實上作的事情是掉換六個數字的位數再相減，不是嗎？#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#我們取任何一個數字ａ#ed_op#SUB#ed_cl#１#ed_op#/SUB#ed_cl#好了；假設原來他在個位數，後來被調整到百位數．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#１００ｘａ#ed_op#SUB#ed_cl#１　#ed_op#/SUB#ed_cl#－　ａ#ed_op#SUB#ed_cl#１#ed_op#/SUB#ed_cl#　＝　９９ｘａ#ed_op#SUB#ed_cl#１#ed_op#/SUB#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#是一個９的倍數．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#再舉另一個數字ａ#ed_op#SUB#ed_cl#５#ed_op#/SUB#ed_cl#好了；假設原來他在萬位數，後來被調整到十位數．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#１０ｘａ#ed_op#SUB#ed_cl#５#ed_op#/SUB#ed_cl#　－　１００００ｘａ#ed_op#SUB#ed_cl#５#ed_op#/SUB#ed_cl#　＝　－９９９０ｘａ#ed_op#SUB#ed_cl#５#ed_op#/SUB#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#仍是９的倍數．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#也就是說，調動任一位數字的位數再予以相減，會是以下的答案：#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#１０#ed_op#SUP#ed_cl#ｙ#ed_op#/SUP#ed_cl#ｘａ#ed_op#SUB#ed_cl#ｘ#ed_op#/SUB#ed_cl#　－　１０#ed_op#SUP#ed_cl#ｘ#ed_op#/SUP#ed_cl#ｘａ#ed_op#SUB#ed_cl#ｘ#ed_op#/SUB#ed_cl#　＝　（１０#ed_op#SUP#ed_cl#ｙ#ed_op#/SUP#ed_cl#－１０#ed_op#SUP#ed_cl#ｘ#ed_op#/SUP#ed_cl#）ｘａ#ed_op#SUB#ed_cl#ｘ#ed_op#/SUB#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#該答案必為９的倍數．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#又由於我們把最大的數字調到最前面，次大的調到第二位．．．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以調整後的新數字絕不會小於原數字，故相減後＂極可能＂大於０．#ed_op#SUB#ed_cl#註一#ed_op#/SUB#ed_cl##ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#HR#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#第二部分：再減去７．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#HR#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#因為大於０又為９的倍數，所以該數減７之後為：#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#（９ｘｍ）－７　＝　（９ｘｎ）＋２，　ｎ≥０#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#故減７之後的答案是一個９的倍數加２．#ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#HR#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#第三部分：六位數字相加．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#HR#ed_cl##ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#９的倍數有一個性質，那就是所有數字相加之後能被９除盡．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#故（９ｘｎ）的每位數字相加的和，是一個９的倍數；#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#而（９ｘｎ）＋２的每位數字相加的和，仍是一個９的倍數加２．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#HR#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#第四部分：相加之後各位數再相加．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#HR#ed_cl##ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#９＝１０－１，　９ｘｎ＝１０ｘｎ－１ｘｎ#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以說若增加２個９；則十位數字增加２，個位數字減少２．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#若增加６個９；則十位數字增加６，個位數字減少６．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#各位數相加的總合仍然不變．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#一個９的倍數加２，可以寫成：#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#（９ｘｎ）＋２＝（１０ｘｎ－１ｘｎ）＋２#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以該數的各位數相加為２．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#故得証．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#註一：#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#由於我們把最大的數字調到最前面，次大的調到第二位．．．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以調整後的新數字絕不會小於原數字，故相減後＂極可能＂大於０．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#但是若六位數字都相同（３３３３３３），則兩數相減為０．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#減去７之後仍然是（９ｘｎ）＋２，但是此時ｎ＝－１；#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#後面的性質就不對了；所以題目減７的部分若是改成加２則會更完美．#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#但是趣味性就減少了．#ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl##ed_op#P#ed_cl#將千元鈔票的6位數字重新排列(由大到小)與原數字相減再檢去7所的數字不斷相加最後得2請問為什麼呀#ed_op#BR#ed_cl#EX 原數字:154111#ed_op#BR#ed_cl##ed_op#BR#ed_cl#541111#ed_op#BR#ed_cl#- 154111#ed_op#BR#ed_cl#=387000#ed_op#BR#ed_cl#- 7#ed_op#BR#ed_cl#=386993=38 =11 =2#ed_op#BR#ed_cl##ed_op#BR#ed_cl#3+8+6+9+9+3=38 3+8=11 1+1=2#ed_op#BR#ed_cl#最後降這6位數相加會等於2 #ed_op#/P#ed_cl##ed_op#/DIV#ed_cl#