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 VerifiedHint: 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: The two digit numbers which are divisible by 3 are-$\Rightarrow$ 12, 15, 18, 21,..................., 99So, 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. |