Question

how many numbers are there between 101 and 999 which are divisible by both 2 and 5

Answer 1

Given that number should be divisible by 2 and 5.

That means should be a multiple of 2 * 5 = 10.

In between 101 and 999 the 1st number divisible by 10 is 110.

Therefore the first term a = 110.

Common difference d = 10.

Last term = 990.

We know that sum of n terms an = a + (n - 1) * d

                                                   990 = 110 + (n - 1) * 10

                                                   990 = 110 + 10n - 10

                                                   990 - 110 = 10n - 10

                                                   880 + 10 = 10n
                        
                                                   890 = 10n

                                                    n = 89.


Therefore the number s between 101 and 999 = 89.


Hope this helps!