Question
How many two-digit numbers are divisible by $3 ?$

Answer

Numbers divisible by $3$ are multiples of $3 .$
$3,6,9,12,15 \ldots$
The smallest two-digit number divisible by $3$ is $12$. And the largest $2-$digit number divisible by $3$ is $99 .$
So, the series of the $2-$digit multiples of $3$ starts with $12$ and ends with $99$ . The difference between the numbers is $3$ .
Therefore, the $A.P.$ is $12,15,18,21,24 \ldots 90,93,96,99$.
In the sequence, the first term, $a=12$.
The last term, $l=99$.
The common difference, $d=3$. The $n^{\text {th }}$ term is $a_n=99$.
Now, $a_n=a_1+(n-1) d$
$\Rightarrow 99=12+(n-1) 3 \Rightarrow 99=12+3 n-3$
$\Rightarrow 99=3 n+9 \Rightarrow 3 n=90$
$\Rightarrow n=30$
Therefore, there are $30$ two$-$digit numbers divisible by $3 .$

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