QUESTION

# How many numbers of two digit are divisible by 3?

Hint: Convert the problem in the form of an AP and use the nth term of an AP formula to determine the number of terms in that sequence.

The two digit numbers which are divisible by 3 are-
$\Rightarrow$ 12, 15, 18, 21,..................., 99
So, this sequence forms an A.P.

First term of the A.P. = ${a_1}$ = 12
Common difference of the A.P. = d = 15 - 12 = 3
Last term of the A.P. = 99

nth term of an A.P. is given by

${a_n} = {a_1} + \left( {n - 1} \right)d$

Substituting the values in the above formula,

$\Rightarrow 99 = 12 + \left( {n - 1} \right)3 \\$
$\Rightarrow \left( {n - 1} \right) = \dfrac{{99 - 12}}{3} = 29 \\$
$\Rightarrow n = 29 + 1 = 30 \\$
Hence there are 30, two digit numbers which are divisible by 3.

Note: These types of problems can be solved using converting the problem statement in the form of a sequence and then use the formulas in that respective sequence to determine the necessary quantities.