Browse · harp Print → jmc prealgebra junior Problem In the multiplication problem below A, B, C, D are different digits. What is A+B? ×CADBCCADD (A) 1 (B) 2 (C) 3 (D) 4 Solution — click to reveal CDCD=CD⋅101, so ABA=101. Therefore, A=1 and B=0, so A+B=1+0=1. Final answer A ← Previous problem Next problem →