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

Last updated date: 15th Mar 2023

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Answer

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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.

Complete step-by-step answer:

$\Rightarrow$ 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

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.

$ \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.

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