Question

How many three digit number are divisible by $7$?

Answer 1

Step 1 : Identify that the sequence is an arithmetic progression

The first three digit number divisible by $7$ is $105$

The last three digit number divisible by $7$ is $994$

The sequence of numbers $105,112,......,994$ which are divisible by $7$ is an arithmetic progression with $1^{st}$ term $105$ and common difference of $7$.

Step 2: Find the number of terms in the given arithmetic progression

The $n^{th}$ term of an arithmetic progression is given by

$a_n=a+(n-1)d$

Here, $a=105,a_n=994,d=7$

$$\begin{gathered} \Rightarrow 994=105+(n-1)7 \\ \Rightarrow 889=7n-7 \\ \Rightarrow 7n=896 \\ \Rightarrow n=128 \end{gathered}$$

Hence, $128$ three digit numbers are divisible by $7$.