Problem: If aaa and bbb are digits for which
2a×b369922989\begin{array}{r} & 2 & a \\ \times & b & 3 \\ \hline & 6 & 9 \\ 9 & 2 & 2 \\ \hline 9 & 8 & 9 & \\ \end{array} ×992b628a3929
then a+b=\mathrm{a}+\mathrm{b}=a+b=
Answer Choices:
A. 333
B. 444
C. 777
D. 999
E. 121212
Solution:
Since a×3\mathrm{a} \times 3a×3 has units digit 999, a must be 333. Hence, b×3\mathrm{b} \times 3b×3 has units digit 222, so b must be 444. Thus, a+b=7\mathrm{a}+\mathrm{b}=7a+b=7.