How many numbers of two digit are divisible by 3?

Solution:

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 .