Question

How many two digit numbers are divisible by 4?

Answer 1

Hint: Here, we will proceed firstly by forming the series of the two-digit numbers which are divisible by 4 and then using the formula for the ${n^{th}}$ term of an AP i.e., ${a_n} = {a_1} + \left( {n - 1} \right)d$ in order to find the number of terms present (n).

Complete step-by-step answer:
As we know that any 2-digit number will have two places (i.e., units place and tens place) which will be occupied by the digits 0,1,2,3,4,5,6,7,8,9.
Clearly, the two digits which are divisible by 4 are given by 12,16,20,……,96.
Since, the common difference between any two consecutive terms in the above series is always 4.
As we know that for any AP with first term as ${a_1}$ and common difference d,
${n^{th}}$ term of the AP, ${a_n} = {a_1} + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$
In the given AP i.e., 12,16,20,……,96, first term is ${a_1} = 12$, common difference is d=4 and ${n^{th}}$ term is ${a_n} = 96$.
Substituting all he values in equation (1), we get
$96 = 12 + 4\left( {n - 1} \right) \\\ \Rightarrow 96 - 12 = 4\left( {n - 1} \right) \\\ \Rightarrow n - 1 = \dfrac{{84}}{4} = 21 \\\ \Rightarrow n = 21 + 1 = 22 \\\$
So, the series of 2-digit numbers which are divisible by 4 consists of a total 22 terms.
Therefore, there are a total 22 two-digit numbers which are divisible by 4.

Note: In this particular problem, if we had to find the sum of all the two digit numbers that are divisible by 4 we will use the formula for the sum of first n terms of any AP i.e., ${{\text{S}}_n} = \dfrac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]$ where n is already been calculated as n=22 and first term is ${a_1} = 12$, common difference is d=4.