Question

Find the number of all the three digit natural numbers which are divisible by 9.

Answer 1

Hint: We know that the lowest three-digit number is 108 and the largest three-digit number is 999. From this we understand that this is an arithmetic progression. So we will use the nth term formula of AP to find the number of terms n that is \[{{T}_{n}}=A+(n-1)d\].

Complete step-by-step answer:
Now the smallest three-digit number that gets divided by 9 is 108. And the largest three-digit number that gets divided by 9 is 999. So the arithmetic progression here is 108, 117, 126,……………999.
So the first term is 108 and the last term is 999. This is an arithmetic progression. So we will use the formula of the nth term of A.P.
\[{{T}_{n}}=A+(n-1)d.......(1)\] where A is the first term and d is the common difference.
So here d is 9 and A is 108 and last term is 999. So substituting all these in equation (1) we get,
\[999=108+(n-1)\times 9.......(2)\]
Now solving for n in equation (2) we get,
\[\begin{align}
  & 999=108+9n-9 \\
 & 9n=999-108+9 \\
 & n=\dfrac{900}{9}=100 \\
\end{align}\]
Therefore, the number of all three digit numbers which are divisible by 9 is 100.

Note: Remembering the nth term formula of the arithmetic progression is the key here. Also in a hurry we can make a mistake in solving equation (2) where we may write 108 in the left hand side of the equation (2) and 999 in the right hand side of the equation (2) and hence we need to be careful while doing this step.