1234567891011121314..........998999
=123456789*10^2880+101112......9899*10^2700+100......998999...(1)
10^2880=(1000)^960=(-1)^960=1(mod1001)=1(mod11)
10^2700=(1000)^900=(-1)^900=1(mod1001)=1(mod11)
101112......99=(10+11++...+99)=109*45=10*45=450=54(mod99)
=10(mod11)
100......998999=(999-998+997-996+........+101-100)=450(mod1001)
=10(mod11)
123456789*10^2880=(1-2+3-4+5-6+7-8+9)*1=5(mod11)
10111213......9899*10^2700=10*1=10(mod11)
100......998999=10(mod11)
代入(1)知1234567891011121314..........998999=5+10+10=25=3(mod11)
所以餘數是3
有獎金嗎?