Question

How many two digit numbers are divisible by 4?
A. 21
B. 22
C. 24
D. 25

Answer 1

Hint: In order to solve this question, first we'll make and A.P. and to find the number of terms we will use the formula of nth term as ${a_n} = a + \left( {n - 1} \right)d$ where a is the first term and d is the common difference.

Complete step-by-step answer:
Numbers divisible by 4 are
$4,8,16,............$
Lowest two digit number divisible by 4 is 12 and the highest two digit number divisible by 4 is 96
So, the series starts with 12 and ends with 96
Difference between the numbers is 4.
So, the A.P. will be $12,16,20,.........96$
We need to find n.
Last term of this series is 96.
Therefore ${a_n} = 96$
First term $a = 12$
And common difference $d = 4$
We know that
${a_n} = a + \left( {n - 1} \right)d$
Substituting the values of a, d and last term
$
   \Rightarrow 96 = 12 + \left( {n - 1} \right) \times 4 \\
   \Rightarrow 96 - 12 = \left( {n - 1} \right) \times 4 \\
   \Rightarrow \left( {n - 1} \right) = \dfrac{{84}}{4} \\
   \Rightarrow n - 1 = 21 \\
   \Rightarrow n = 22 \\
$

Therefore, there are 22 two digit numbers divisible by 4.
Hence, the correct option is “B”.

Note: In order to solve these types of problems, first of all remember the formula of arithmetic progression. Remember how to find the nth term of and A.P. Also remember how to find the sum of n terms of an A.P. Similarly learn about geometric progression and harmonic progression. This will help a lot to solve problems related to A.P, G.P and H.P. The above question is an application of A.P.