The sum of all two-digit numbers divisible by 3 is:
Answer: A
The two-digit numbers divisible by 3 are 12, 15, ..., 99. This is an A.P. with \(a=12, d=3, a_n=99\). First, find the number of terms: \(a_n = a+(n-1)d \Rightarrow 99 = 12+(n-1)3 \Rightarrow 87 = (n-1)3 \Rightarrow 29 = n-1 \Rightarrow n=30\). The sum is \(S_n = \frac{n}{2}(a+a_n) = \frac{30}{2}(12+99) = 15(111) = 1665\).