If the number 6a34b4 is divisible by 3. How many values of a + b are possible?
3
1
4
2
Here, 6 + a + 3 + 4 + b + 4 = 17 + a + b
Now, a + b cannot be more than 9.
Thus, only a + b = 1, 4, 7 satisfy the divisibility by 3.