Question

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

Answer 1

Hint:
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.