Problem:
If a,b, and c are digits for which
−74ca872b3
then a+b+c=
Answer Choices:
A. 14
B. 15
C. 16
D. 17
E. 18
Solution:
The subtraction problem posed is equivalent to the addition problem
+ 4c7872b32
which is easier to solve. Since b+3=12, b must be 9. Since 1+8+7 has units digit a, a must be 6. Because 1+4+c=7,c=2. Hence a+b+c =6+9+2=17.