Problem: The three-digit number 2a32a32a3 is added to the number 326326326 to give the three-digit number 5b95 b 95b9. If 5b95b95b9 is divisible by 999, then a+b\mathrm{a}+\mathrm{b}a+b equals:
Answer Choices:
A. 222
B. 444
C. 666
D. 888
E. 999 Solution:
Since 5b95 b 95b9 is divisible by 9, b=4∴a+2=4,a=29, \mathrm{~b}=4 \quad \therefore \mathrm{a}+2=4, \mathrm{a}=29, b=4∴a+2=4,a=2, and a+b=6\mathrm{a}+\mathrm{b}=6a+b=6.