How many numbers of two digit are divisible by 3?

In this problem, we need to find out how many numbers of two digits are divisible by 3.

So, we know that the first two digit number that is divisible by 3 is 12 and the last two digit number divisible by 3 is 99. Also, all the terms which are divisible by 3 will form an A.P. with the common difference of 3.

So here,

First term (a) = 12

Last term (an) = 99

Common difference (d) = 3

So, let us take the number of terms as n

Now, as we know,

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

So, for the last term,

$99=12+(n-1) 3$

$99=12+3 n-3$

$99=9+3 n$

$99-9=3 n$

Further simplifying,

$90=3 n$

$n=\frac{90}{3}$

$n=30$

Therefore, the number of two digit terms divisible by 3 is 30 .