All 2-digit numbers divisible by 6 are as follows: –

6, 12, ………., 96

The above series of numbers forms an arithmetic progression with

first term(a) = 6 and,

common difference(d) = (n + 1)th term – nth term = 12 – 6 = 6

last term or nth term(a_n) = 96

Let the number of terms in above series be n.

∵ a_n = a + (n – 1) × d

⇒ 96 = 6 + (n – 1) × 6

⇒ 90 = 6n – 6

⇒ 6n = 96

∴ n = 16

Thus, total no. of all 2-digit numbers divisible by 6.