由 piny 於 星期三 三月 15, 2006 2:13 pm
#ed_op#DIV#ed_cl#如果第二位數是指該門牌號碼的第二個數字的話...#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#那應該是64號#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#推理如下#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#先用實際情況推演,由於是否小於500和是否為平方數,會剛好與瓊斯聽到的相反,但是否為立方數則為屬實。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#以(是,是,是)來形容是否小於500、是否為平方數、是否為立方數。(瓊聽到的,但以實際情況來推算)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1.(是,是,是)→512,1000#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2.(是,否,是)→729#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#3.(是,是,否)→大於等於500的數,扣掉1,2,4之數#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#4.(是,否,否)→大於500的平方數,扣掉729#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#5.(否,是,是)→27,125,216,343#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#6.(否,否,是)→64#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#7.(否,是,否)→大於等於13小於500的數,扣掉5,6,8之數#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#8.(否,否,否)→大於等於13小於500的平方數,扣掉64#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#可看出如果瓊聽到這些選項,而要求要曉得第二位數是否為1,則可推論其所剩可能必為兩個,一個是第二位為1,另一個則否。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#因此可看出(是,是,是)之實際情況符合,而此情況在瓊所聽到的狀況會成為(否,否,是),瓊只會由是否第二位為1來猜512或1000,當然都猜錯囉,而(否,否,是)之實際情況只有64,故64為所求。#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#