Question

How many two digit numbers are divisible by $7$?

Answer 1

Step 1 : Find the arithmetic progression (A.P.) from the given information.

The smallest and largest two-digit numbers that are divisible by seven are $14\text{and}98$.

The common difference will be $7$ because the numbers are divisible by seven.

Therefore, the first and last terms of the AP $14,21,28,\dots ,98$ are $14\text{and}98$.

Step 2 : Calculating the required number of terms,

We know that, $\mathit{l}\mathbf{=}\mathit{a}\mathbf{+}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{d}$, where $l$ is the last term, $a$ is the first term, $n$ is the number of terms, and $d$ is a common difference.

Here, $l=98,a=14,\text{and}d=7.$

$$\begin{gathered} \therefore l=a+\left(n-1\right)d \\ \Rightarrow 98=14+\left(n-1\right)7 \\ \Rightarrow 98=7n+7 \\ \Rightarrow n=13 \end{gathered}$$

Final Answer :

Hence, there are $13$ two digit numbers that are divisible by $7$.