#ed_op#P#ed_cl#
宇智波鼬 寫到:piny 寫到:<DIV>1735+3465n(n為非負整數)</DIV><DIV>1736+3465n(n為非負整數)</DIV><DIV>1737+3465n(n為非負整數)</DIV><DIV>1738+3465n(n為非負整數)</DIV>
#ed_op#BR#ed_cl##ed_op#BR#ed_cl#請問是如何求得的? (好像用的方法和我類似...)
#ed_op#/P#ed_cl##ed_op#P#ed_cl#與韓信點兵的方法類似#ed_op#/P#ed_cl##ed_op#P#ed_cl#先看5和7#ed_op#/P#ed_cl##ed_op#P#ed_cl#7,14,21,28,35,42,49,...(就是看七的倍數)#ed_op#BR#ed_cl#6,13,20,27,36,41,48,...(上述數列減1)#ed_op#/P#ed_cl##ed_op#P#ed_cl#可看出20可被5整除#ed_op#BR#ed_cl#故符合5與7者為20+35n(n為非負整數)....(1)#ed_op#/P#ed_cl##ed_op#P#ed_cl#再看9和11#ed_op#/P#ed_cl##ed_op#P#ed_cl#11,22,33,44,55,66,...(11的倍數)#ed_op#BR#ed_cl#10,21,32,43,54,65,...(上述數列減1)#ed_op#/P#ed_cl##ed_op#P#ed_cl#可看出54可被9整除#ed_op#BR#ed_cl#故符合9與11者為52+99m(m為非負整數)....(2)#ed_op#/P#ed_cl##ed_op#P#ed_cl#由(1)(2)依同樣方法再做一次可得答案#ed_op#/P#ed_cl##ed_op#P#ed_cl# #ed_op#/P#ed_cl#