Q150 我想是 B, 如錯, 請見諒!!! 2tu = 3a (a for any intergers) 2tu*3 = 3a*3 = 9a 2tu*3=(200+10t+u)*3=9a 600+30t+3u=9a (594+6)+(27t+3t)+3u=9a (594+27t)+(6+3t+3u)=9a 9(66+3t)+3(2+t+u)=9a Since the former part can be divisble by 9, so...the latter part can also be divisble by 9(without any reminders) 3(2+t+u), 3 is not divisble by 9 , so 2+t+u must be divisble by 9! -- by 会员 treewing929 (2012/2/7 21:15:30)
你这个方法也很不错呀!但是容易做到后面出错~我的理解是,做到这一步:9(66+3t)+3(2+t+u)=9a 加号的前后两个部分都应该可以被9整除,所以3(2+t+u)应该可以被9整除。但是不能确定2+t+u就能被9整除,因为可能2+t+u正好只是3的倍数,和括号前的3相乘正好可以被9整除,这样的话它只能被3整除;2+t+u也可能正好是9的倍数,这样子就不充分了。不知道我的思路有没有错。 |