We have to find the numbers between 101 and 999 which are divisible by both 2 and 5

That means the numbers divisible by 10 between 101 and 999.

Let, the required divisible numbers be,

110, 120, 130, …, 990

a=110, d=10, last term, l=990

Number of terms, n = +1

= + 1

= +1

=89

Hence, the number of natural numbers between 101 and 999 which are divisible by both 2 and 5 are 89.