How many two digit numbers are not divisible by 3

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 $\lfloor\frac{999}{3*5*11}\rfloor - \lfloor\frac{99}{3*5*11}\rfloor = 6$

Answer

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:

So, this sequence forms an A.P.

First term of the A.P. = ${a_1}$ = 12Common 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.

How many two-digits numbers are divisible by 3

Solution;

The two-digit numbers divisible by 3 are 12, 15, 18, …, 99. Clearly, these number are in AP.

Here, a = 12 and d =15 —12 = 3

Let this AP contains n terms. Then,

an = 99

12 + (n – 1) x 3=99         [an=a+(n —1)cl]

3n-F9=99

3n = 99-9 = 90

n = 30

Hence, there are 30 two-digit numbers divisible by 3.