How many numbers of two digit are divisible by 3?
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.
Last updated date: 20th Sep 2023
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