Question

How many two digit numbers are divisible by $13$?

Answer 1

The first two digit number divisible by $13$ is $13$, the next two digit number divisible by $13$ is $26$ and next is $39$ and so on. The last two digit number divisible by $13$ is $91$. Therefore, the required sequence is:

$13,26,39,.......91$

From the above sequence, we get that the first term is $a_1=13,a_2=26$ and the $n^{th}$ term is $T_n=91$

Now, we find the common difference as follows:

$d=a_2-a_1=26-13=13$

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$

To find the number of terms of the A.P, substitute $a=13$, $T_n=91$ and $d=6$ in $T_n=a+(n-1)d$ as follows:

$T_n=a+(n-1)d\Rightarrow 91=13+(n-1)13\Rightarrow 91=13+13n-13\Rightarrow 13n=91\Rightarrow n=\frac{91}{13}\Rightarrow n=7$

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