The total numbers divisible by $7$ below $100$ would be given by $\left \lfloor \dfrac{100}{7} \right \rfloor = 14$ . But one of them is $7$ which is one-digit number. Thus the number of required two digit numbers $=14-1 = \large\boxed{13}$

The very first $2$ digit multiple of $7$ is $14$ and the $98$ is the last. The only thing we have to do is to find the number of terms of this arithmetic progression.

$a = 14$ , $d= 7$

${ a }_{ n } = a + (n-1)d$

$\Rightarrow 98 = 14 + (n-1)7$

$\Rightarrow n= 13$

Divide by 7 with highest divisible no. of 7 and then subtracted by 1

In the table of 7 the highest no by which the answer is in two digit excluding 1 because 7 is one digit So the answer is 13

As we know, 98 is the last two digits no divisible by 7 98/7 = 14 but in these 14 numbers we have also counted 7 which is of one digit i.e. we will substract this case from 14, 14 - 1 = 13

Why aren't we including negatives? This would count all two-digit numbers between -99 and 99, meaning the solution should be 26.