Question

How many numbers are there between 101 and 999, which are divisible by both 2 and 5?

Answer 1

Answer:

Step-by-step explanation:

Given :-

101 and 999 are natural numbers.

To Find :-

Number of terms which are divisible by both 2 and 5.

Formula to be used :-

$\boxed{\bf { a_n=a+(n-1)\times d}}$

Solution :-

The sequence goes like this,

110, 120, 130, ..........., 990

Since, they have the common difference of 10, they form an A.P.

a = 110, a(n) = 990 and d = 10

Putting all the values, we get

$\sf \implies a_n=a+(n-1)\times d}$

$\sf \implies 990=110+(n-1)\times 10}$

$\sf \implies 990-110=(n-1)\times10$

$\sf \implies 880=(n-1)\times 10$

$\sf \implies n-1 = 88$

$\bf \implies n=89$

Hence, there are 89 terms between 101 and 999 which are divisible by 2 and 5.