1999…(2002个9)+999…99(2002个9)×999…9(2002个9)的末尾有多少个零

令a=2009个0
则199（2009个9)
=100 (2009个0)+a
=(a+1)+a
=2a+1
所以原式=2a+1+a²
=(a+1)²
=[100 (2009个0)]²
=100 (2009×2个0)
即4018个0额，是2002个9哦，对不起令a=2002个0则199（2002个9)=100 (2002个0)+a=(a+1)+a=2a+1所以原式=2a+1+a²=(a+1)²=[100 (2002个0)]²=100 (2002×2个0)即4004个0