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**Hint**: First of all find the positive integers less than or equal to 100 which are divisible by 2, 3 and 5 which we are going to do by writing the integers which are separately divisible by 2, 3 and 5. Now, total of these integers contain numbers which are repeating for e.g., the integers which are divisible by 2 also contain integers which are divisible by 3 and 5 also so we are going to subtract the positive integers which are divisible by 6, 15 and 10 from the total integers which are divisible by 2, 3 and 5. Now, we have subtracted extra 3 integers which are divisible by 2, 3 and 5 so we have to add 3 integers to the result of subtraction. Now, we have got the integers which are divisible by 2, 3 and 5 so subtracting these many integers from 100.

**:**

__Complete step-by-step answer__We are asked to find the positive integers till 100 which are not divisible by 2, 3 or 5.

First of all we are going to find the positive integers till 100 which are divisible by 2, 3, 5, 6, 10, 15, 30.

The positive integers till 100 which are divisible by 2 are:

2, 4, 6, 8 ,10, 12, 14………….100

The above series is in A.P. in which the first term is 2 and common difference is 2.

We know that the general term of an A.P. is equal to:

$ {{T}_{n}}=a+\left( n-1 \right)d $ …………. Eq. (1)

Substituting $ {{T}_{n}}=100 $ , “a” and “d” as 2 and 2 respectively in the above formula we get,

$ \begin{align}

& 100=2+\left( n-1 \right)\left( 2 \right) \\

& \Rightarrow 100=2+2n-2 \\

& \Rightarrow 100=2n \\

\end{align} $

Dividing 2 on both the sides we get,

$ n=50 $

50 numbers of positive integers till 100 are divisible by 2.

Now, the number of positive integers which are divisible by 3 till 100 are:

3, 6, 9, 12, 15……….99

The above series is an A.P. with first term (a) as 3 and common difference as (d) as 3.

$ {{T}_{n}}=a+\left( n-1 \right)d $

Substituting $ {{T}_{n}}=99 $ , “a” and “d” as 3 and 3 respectively in the above formula we get,

$ \begin{align}

& 99=3+\left( n-1 \right)\left( 3 \right) \\

& \Rightarrow 99=3+3n-3 \\

& \Rightarrow 99=3n \\

\end{align} $

Dividing 3 on both the sides we get,

$ 33=n $

33 positive integers till 100 are divisible by 3.

Number of positive integers which are divisible by 5 till 100 are:

5, 10, 15, 20………..100

In the above series, first term (a) is 5 and common difference (d) is 5 so substituting the value of “a” as 5 and “d” as 5 and $ {{T}_{n}} $ as 100 in eq. (1) we get,

$ \begin{align}

& {{T}_{n}}=a+\left( n-1 \right)d \\

& \Rightarrow 100=5+\left( n-1 \right)5 \\

& \Rightarrow 100=5+5n-5 \\

& \Rightarrow 100=5n \\

\end{align} $

Dividing 5 on both the sides we get,

$ \begin{align}

& \dfrac{100}{5}=n \\

& \Rightarrow 20=n \\

\end{align} $

Now, the above number of positive integers contains numbers which are divisible by 2 and 3 also so we have to subtract these many integers from 20.

Adding the integers which are divisible by 2, 3 and 5 we get,

$ \begin{align}

& 50+33+20 \\

& =103 \\

\end{align} $

The positive integers which are divisible by 6 are:

6, 12, 18, 24,………96

In the above series, first term (a) is 6 and common difference (d) is 6 so substituting the value of “a” as 10 and “d” as 10 and $ {{T}_{n}} $ as 100 in eq. (1) we get,

$ \begin{align}

& {{T}_{n}}=a+\left( n-1 \right)d \\

& \Rightarrow 96=6+\left( n-1 \right)6 \\

& \Rightarrow 96=6+6n-6 \\

& \Rightarrow 96=6n \\

\end{align} $

Dividing 6 on both the sides of the above equation we get,

$ \begin{align}

& \dfrac{96}{6}=n \\

& \Rightarrow n=16 \\

\end{align} $

16 numbers of positive integers are divisible by 6.

The positive integers which are divisible by 10 are:

10, 20, 30………100

In the above series, first term (a) is 10 and common difference (d) is 10 so substituting the value of “a” as 10 and “d” as 10 and $ {{T}_{n}} $ as 100 in eq. (1) we get,

$ \begin{align}

& {{T}_{n}}=a+\left( n-1 \right)d \\

& \Rightarrow 100=10+\left( n-1 \right)10 \\

& \Rightarrow 100=10+10n-10 \\

& \Rightarrow 100=10n \\

\end{align} $

Dividing 10 on both the sides of the above equation we get,

$ \begin{align}

& \dfrac{100}{10}=n \\

& \Rightarrow n=10 \\

\end{align} $

10 positive integers are divisible by 10.

The positive integers till 100 which are divisible by 15 are:

15, 30, 45,……..90

In the above series, first term (a) is 15 and common difference (d) is 15 so substituting the value of “a” as 15 and “d” as 15 and $ {{T}_{n}} $ as 90 in eq. (1) we get,

$ \begin{align}

& {{T}_{n}}=a+\left( n-1 \right)d \\

& \Rightarrow 90=15+\left( n-1 \right)15 \\

& \Rightarrow 90=15+15n-15 \\

& \Rightarrow 90=15n \\

\end{align} $

Dividing 15 on both the sides we get,

$ \begin{align}

& \dfrac{90}{15}=n \\

& \Rightarrow 6=n \\

\end{align} $

6 positive integers are divisible by 15.

Adding the positive integers which are divisible by 6, 15, 10 we get,

$ \begin{align}

& 16+10+6 \\

& =32 \\

\end{align} $

Now, subtracting 32 from 103 we get,

$ \begin{align}

& 103-32 \\

& =71 \\

\end{align} $

The positive integers till 100 which are divisible by 30 are:

30, 60, 90

These 3 positive integers are extra subtracted from the latest subtraction that we did so adding 3 in 71 we get,

$ \begin{align}

& 71+3 \\

& =74 \\

\end{align} $

Now, to get the positive integers which are not divisible by 2, 3 and 5 and are less than or equal to 100 is calculated by subtracting 74 from 100 we get,

$ \begin{align}

& 100-74 \\

& =26 \\

\end{align} $

**So, the correct answer is “Option B”.**

**Note**: The possible mistake that could happen in the above problem is that either you forgot to add 3 in the subtraction of total integers which are divisible by 2, 3 and 5 with the integers which are divisible by 6, 10, 15 or you instead of adding 3 you might have subtracted 3 from the subtraction result.

Then either the number of positive integers which are divisible by 2, 3 or 5 when you forgot to add 3 is equal to:

$ \begin{align}

& 103-\left( 16+10+6 \right) \\

& =103-32 \\

& =71 \\

\end{align} $

And instead of adding 3, subtracting 3 from 71 we get,

$ \begin{align}

& 71-3 \\

& =68 \\

\end{align} $

These are the mistakes that could happen.

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