Question

How many numbers of two digits are divisible by $7$?

Answer 1

We observe that $14$ is the first two-digit number divisible by $7$ and $98$ is the last two-digit number divisible by $7$.

Thus, we have to determine the number of terms in the sequence.
$14,21,28,...,98$

Clearly, it is an A.P. with first term $=14$ and common difference $=7$ i.e. $a=14$ and $d=7$.

Let this be the $n^{th}$ term in this A.P.

Then, $n^{th}$ term $=98$

$⟹14+(n-1)\times 7=98$
$⟹14+7n-7=98$
$⟹7n=91⟹n=13$
Hence, there are $13$ numbers of two digits which are divisible by $7$.

Alternatively,

The two digit numbers are from $10$ to $99$

When we divide $10$ by $7$, we get the quotient as $1$. $[10=7\times 1+3]$

Also, when we divide $99$ by $7$, we get the quotient as $14$. $[99=7\times 14+1]$

So, the two digit numbers that are divisible by $7$ are: $14-1=13$