Question

# How many numbers between $1$ and $200$ are multiples of $5$ and divisible by $3$?A. $100$B. $50$C. $95$D. $93$

Hint: Indirectly we need to calculate the terms that are multiples of 5 plus the terms that divisible by 3 minus the terms that are divisible by ${\text{15}}$, as there will be few common terms in both the series. Here we use the concept of AP to get the number of terms.

As per the given data, we have to find the between $1$ and $200$ that are multiples of $5$ and should also be divisible by $3$
The number that is multiple of $5$ between $1$ and $200$ is $5,10,15...200$ and number divisible by $3$ are $3,6,9...198$
The number satisfying both the conditions are $15,30,...195$
In the above-obtained series, we can observe that the first term is $15$ and the difference is also $15$ while the last term is $195$.
Now, substituting the values in the formula of A.P, i.e.
${{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}}$ we get,
${\text{195 = 15 + }}\left( {{\text{n - 1}}} \right){\text{15}} \\ \Rightarrow {\text{180 = (n - 1)15}} \\$
Simplifying it further and calculating the value of n, we get,
$\Rightarrow \dfrac{{180}}{{15}}{\text{ = n - 1}}$
On simplifying further we get,
$\Rightarrow {\text{12 = n - 1}}$
On rearranging we get,
$\Rightarrow {\text{12 + 1 = n}}$
Hence , ${\text{n = 13}}$
Hence , In the above obtained series we can observe that first term is ${\text{5}}$ and difference is also ${\text{5}}$ while the last term is $195$
Now, substituting the values in the formula of A.P, i.e.
${{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}}$ we get,
$\Rightarrow {\text{195 = 5 + }}\left( {{\text{n - 1}}} \right){\text{5}} \\ \Rightarrow {\text{190 = (n - 1)5}} \\$
On calculating the value of n we get,
$\Rightarrow \dfrac{{190}}{5}{\text{ = n - 1}}$
On simplifying we get,
$\Rightarrow {\text{38 = n - 1}}$
$\Rightarrow {\text{ n = 39}}$
Now, calculating the total number of terms that are divisible by $3$.
Hence, In the above-obtained series we can observe that first term is ${\text{3}}$ and difference is also ${\text{3}}$ while the last term is $198$
Now, substituting the values in the formula of A.P, i.e.
${{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}}$ we get,
$\Rightarrow {\text{198 = 3 + }}\left( {{\text{n - 1}}} \right)3 \\ \Rightarrow {\text{195 = (n - 1)3}} \\$
Hence calculating the value of n,
$\Rightarrow \dfrac{{195}}{3}{\text{ = n - 1}}$
On further simplification we get,
$\Rightarrow {\text{66 = n - 1}}$
$\Rightarrow {\text{n = 67}}$
Hence , we need to add all the terms satisfying the condition of divisible by $3$ and multiples of $5$ and subtract the terms common in both of them
So, it will be
${\text{67 + 39 - 13 = 93}}$
Hence, option (d) is our correct answer.

Note: Multiples of a number like here we take 5, are always divisible by 5. So we can say that if a number is multiple of 5 it is divisible by 5. Also here we subtract the numbers divisible by 15 from the list, as we have already considered them in the list of numbers divisible by 3 and in the list of numbers which are multiples of 5, so to count them only once we subtract the total sum, and hence our required answer is obtained.