2003 AMC 12B Problems/Problem 18

suppose n=a*10^4+b*10^3+c*10^2+d*10+e, where a,b,c,d,e are integers between [0,9] then q=(a*10^2+b*10+c)%11, r=d*10+e, and n%11=(a*100+b*10+c)%11 (q+r)%11 = n%11, since 10000<=n<=99999 there are 9090-910+1=8181 n values that are multiples of 11, thus there are 8181 (q+r) values that are multiples of 11