Question

How many 3-digit numbers are divisible by 9?

Answer 1

Hint: Here, we will find the smallest and largest 3-digit number divisible by 9. We will use the formula for the \[{n^{th}}\] term of an arithmetic progression. Then, we will take the required smallest 3-digit number as the first term, the required largest 3-digit number as the last term and 9 as the common difference of the A.P. We will find the number of terms of the A.P. and the number of terms will be the number of 3-digit numbers divisible by 9.

Formulas used: We will use the formula, \[l = a + \left( {n - 1} \right)d\] where \[l\] is the last term, \[d\] is the common difference and \[n\] is the total number of terms in an A.P.

Complete step-by-step answer:
We know that the divisibility rule of 9 is given as: if the sum of digits of a number is divisible by 9, the number is also divisible by 9.
Let’s go through the multiples of 9 and find the smallest 3-digit number that is a multiple of 9.
We know that
\[\begin{array}{l}9 \times 11 = 99\\9 \times 12 = 108\end{array}\]
We have found that 108 is the smallest 3-digit number that is divisible by 9.
Now we will find the largest 3-digit number that is a multiple of 9.
The largest 3 digit number is 999, we can see that 999 is divisible by 9 because it satisfies the divisibility rule for 9. The sum of digits of 999 is 27 and 27 is divisible by 9.
Let us substitute 108 for \[a\], 999 for \[l\] and 9 for \[d\] in the formula \[l = a + \left( {n - 1} \right)d\].
\[999 = 108 + \left( {n - 1} \right)9\]
Subtract 108 from both sides.
\[\begin{array}{l} \Rightarrow 999 - 108 = 108 + \left( {n - 1} \right)9 - 108\\ \Rightarrow 891 = \left( {n - 1} \right)9\end{array}\]
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 to both sides, we get
\[ \Rightarrow 99 + 1 = n - 1 + 1\\
\Rightarrow 100 = n\]
We have found that there are 100 terms.
$\therefore $ The number of 3-digit numbers divisible by 9 is 100.

Note: We have used the method of arithmetic progression to solve this question. Arithmetic progression is a type of sequence or series of numbers in which the difference between any two numbers of that series is constant. That means the common difference between numbers is the same for a particular series. We have also used the divisibility rule here because it makes the process much easier to find if a number is divisible by another number without performing the actual division.