Question

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5

Answer 1

since the number must be divisible by both 2 and 5 ,

first term, a = 110 common difference, d = 10 last term , l = 990

therefore, the number of terms is

n= [last term - first term / diffrence]+1

= [990 - 110 / 10] + 1

= 88+1 = 89

therefore n = 89

therefore there are 89 terms that are divisible by 2 ,5