Question

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

Answer 1

Here, we will use the concept of arithmetic progression to solve the question. The number of three-digit natural numbers divisible by 9 can be written in the form of an A.P. We will use the formula for ${n^{{\text{th}}}}$ term of an A.P. and simplify the equation to get the number of terms in the A.P. and hence, the number of three-digit natural numbers divisible by 9.
Formula Used: The ${n^{{\text{th}}}}$ term of an A.P. is given by the formula ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term of the A.P.,
$d$ is the common difference, and $n$ is the number of terms in the A.P.

Complete step by step solution:
We will use the formula for ${n^{{\text{th}}}}$ term of an arithmetic progression to find the number of three-digit natural numbers divisible by 9.
Let the number of three-digit natural numbers which are divisible by 9 be $n$.
We know that the first three digit natural number is 100 and the last three digit natural number is 999.
The first three digit natural number divisible by 9 is 108.
The second three digit natural number divisible by 9 is $108 + 9 = 117$.
The third three digit natural number divisible by 9 is $117 + 9 = 126$.
Also, we can observe that 999 is also divisible by 9.
Therefore, we get the sequence of three digit natural numbers divisible by 9 as
108, 117, 126, ……, 999
We can observe that the sequence forms an arithmetic progression with first term 108, common difference 9, and the last term 999.
Now, we know that the ${n^{{\text{th}}}}$ term of an A.P. is given by the formula ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term of the A.P., $d$ is the common difference, and $n$ is the number of terms in A.P.
Substituting $a = 108$, $d = 9$, ${a_n} = 999$ and in the formula, we get
$999 = 108 + \left( {n - 1} \right)9$
We will simplify this equation to get the number of terms in the A.P.
Subtracting 108 from both sides of the equation, we get
$\begin{array}{l} \Rightarrow 999 - 108 = 108 + \left( {n - 1} \right)9 - 108\\\ \Rightarrow 891 = \left( {n - 1} \right)9\end{array}$
Dividing both sides by 9, we get
$\begin{array}{l} \Rightarrow \dfrac{{891}}{9} = \dfrac{{\left( {n - 1} \right)9}}{9}\\\ \Rightarrow 99 = n - 1\end{array}$
Adding 1 on both sides, we get
$\begin{array}{l} \Rightarrow 99 + 1 = n - 1 + 1\\\ \Rightarrow 100 = n\end{array}$
Thus, the number of terms in the A.P. 108, 117, 126, ……, 999 is 100.

Therefore, the number of three-digit natural numbers divisible by 9 is 100.

Note:
We have used the concept of A.P. to solve the given question. An arithmetic progression is a sequence where the difference of every two consecutive numbers is a fixed constant. We have to remember the formula for ${n^{{\text{th}}}}$ term of an A.P. A common mistake we can make is to use $n$ instead of $n - 1$ and leave the answer as 99 three-digit natural numbers. This will give us the wrong answer.