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!
if you understand this,no problem 一个数是否能够被4整除，只要看它的后两位。 一个数是否能够被8整除，只要看它的后三位。 一个数能否被3或9整除，取决于各位之和能否被3整除。 一个数能否被6整除，分别考虑被2，和被3除时的情况 一个数能否被11整除，错位相加再相减。 the correct answer is E
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!
if you understand this,no problem 一个数是否能够被4整除，只要看它的后两位。 一个数是否能够被8整除，只要看它的后三位。 一个数能否被3或9整除，取决于各位之和能否被3整除。 一个数能否被6整除，分别考虑被2，和被3除时的情况 一个数能否被11整除，错位相加再相减。 the correct answer is E
京公网安备11010202008513号 京ICP证101109号 京ICP备12012021号
发表于 2012-2-7 12:39:01
