Number of zeros at then end of 991001+1 is

Let n=1001, we can write991001+1=(100−1)n+1                 =nC0100n−nC1100n−1+nC2100n−2−…                  =100m+nCn−1(100)−1+1where      m=nC0100n−1−nC1100n−2+nC2100n−3−…−nCn−2100+nCn−1As  nCn−1=1001,the units digit of m is different from 0. Thus, the number of zeros at the end of 991001+1is two.