Answer
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.
A simple approach I like is the Inclusion-exclusion principle.
Let $C_{[i1, i2,...]}$ mean number of three digit numbers divisible by i1, i2, ... and $C$ the number of three digit numbers. So we have:
$C - C_{[3]} - C_{[5]} - C_{[11]} + C_{[3, 5]} + C_{[3, 11]} + C_{[5, 11]} - C_{[3, 5, 11]}$
For each $C_{[i_1, i_2, ...]}$ we have:
$C_{[i_1, i_2, ...]} = \lfloor\frac{999}{i_1 * i_2 *...}\rfloor - \lfloor\frac{99}{i_1 * i_2 *...}\rfloor$
This results in $900 - (300 + 180 + 81) + (60 + 27 + 17) - 6 = 437$
Note:
I assume you mean "How many three digit numbers are not divisible by 3,5 or 11?"
Otherwise your answer is $\lfloor\frac{999}{3*5*11}\rfloor - \lfloor\frac{99}{3*5*11}\rfloor = 6$