The numbers between 101 and 999 that are divisible by both 2 and 5 are 110, 120, 130,…, 990.

This forms an AP with first term a = 110

and common difference = d = 10

Last term is 990.

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

990 = a + (n - 1)d

⇒ 990 = 110 + (n - 1)10

⇒ 990 - 110 = 10n - 10

⇒ 880 + 10 = 10n

⇒ 890 = 10n

⇒ n = 89

There are 89 numbers between 101 and 999 that are divisible by both 2 and 5.