The three digit numbers divisible by 9 are 108, 117, 126, …., 999.

This forms an AP with first term a = 108

and common difference = d = 9

Last term is 999.

Now, number of terms in this AP are given as:

999 = a + (n - 1)d

⇒ 999 = 108 + (n - 1)9

⇒ 999 - 108 = 9n - 9

⇒ 891 + 9 = 9n

⇒ 900 = 9n

⇒ n = 100

There are 100 three - digit numbers that are divisible by 9.