Question

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

Answer 1

Hint: Make a series which will be AP whose first term is 12 and last term is 99 taking the common difference as 3.

Complete step-by-step answer:
We know, first two digit number divisible by $3$ is $12$ and the last two digit number divisible by $3$ is $99$.

Thus, we get \[12,15,18,...,99\;\] which is an AP

Here, a = $12$ and d = $3$ are first Term & common difference.
Let there be n terms. Then,
We know the last two digit number is $99$ in the series , therefore ${a_n}$= 99 it is also called the nth term or the last term since there are n terms in the series therefore the nth term will be the last term. Here we have to find the number of terms.
So we can write,

\[{{a_n} = 99} \\
  \Rightarrow {a + (n - 1)d = 99} \\
  \Rightarrow {12 + (n - 1)3 = 99} \\
  \Rightarrow {n = 29 + 1 = 30}
\]

Therefore, there are 30 two digit numbers divisible by 3.

Note: In these types of questions we should always try to make a series. It may be an AP or a GP. After making the series solve the portion from which you can get what you have been asked, for an example we have to find here a number of terms so that we get n.