Question

Find the number of positive integers from 1 to 1000 which are divisible by at least 2, 3 or 5.

Answer 1

Hint: In this problem, we have to find the number of positive integers from 1 to 1000 which are divisible by at least 2, 3 or 5. We can first assume the numbers divisible by 2 as A, the number divisible by 3 as B and the number divisible by 5 as C. We can then find their unions and intersections. We can then use the formula,\[n\left( A\cup B\cup C \right)=n\left( A \right)+n\left( B \right)+n\left( c \right)-n\left( A\cap B \right)-n\left( B\cap C \right)-n\left( C\cap A \right)+n\left( A\cap B\cap C \right)\] to find the number of integers divisible by 2, 3 or 5.

Complete step by step solution:
Here we have to find the number of positive integers from 1 to 1000 which are divisible by at least 2, 3 or 5.
We can now assume,
\[n\left( A \right)\] as the number divisible by 2
\[n\left( B \right)\] as the number divisible by 3
\[n\left( C \right)\] as the number divisible by 5.
We have to find the numbers between 1 to 1000, so we can write each of the divisible numbers as
\[\begin{align}
  & \Rightarrow n\left( A \right)=\dfrac{1000}{2}=500.....(1) \\
 & \Rightarrow n\left( B \right)=\dfrac{1000}{3}=333......(2) \\
 & \Rightarrow n\left( C \right)=\dfrac{1000}{5}=200.......(3) \\
\end{align}\]
We can now find the intersections, we get
\[\begin{align}
  & \Rightarrow n\left( A\cap B \right)=\dfrac{1000}{2\times 3}=\dfrac{1000}{6}=166......(4) \\
 & \Rightarrow n\left( B\cap C \right)=\dfrac{1000}{3\times 5}=\dfrac{1000}{15}=66.......(5) \\
 & \Rightarrow n\left( C\cap A \right)=\dfrac{1000}{5\times 2}=\dfrac{1000}{10}=100......(6) \\
 & \Rightarrow n\left( A\cap B\cap C \right)=\dfrac{1000}{2\times 3\times 5}=\dfrac{1000}{30}=33.....(7) \\
\end{align}\]
We know that to find the number of integers divisible by 2, 3 or 5, we use the formula
\[n\left( A\cup B\cup C \right)=n\left( A \right)+n\left( B \right)+n\left( c \right)-n\left( A\cap B \right)-n\left( B\cap C \right)-n\left( C\cap A \right)+n\left( A\cap B\cap C \right)\]
We can now substitute the values (1), (2), (3), (4), (5), (6), (7) in the above formula, we get
\[\Rightarrow n\left( A\cup B\cup C \right)=500+33+200-166-66-100+33=734\]
Therefore, there are 734 positive integers which are divisible by at least 2, 3 or 5 from 1 to 1000.

Note: We should always remember that if we divide a fraction, we may get a decimal number as the answer, but here we have to find the number of terms, so we can round it off and write the answer for such steps. We should also remember the formula \[n\left( A\cup B\cup C \right)=n\left( A \right)+n\left( B \right)+n\left( c \right)-n\left( A\cap B \right)-n\left( B\cap C \right)-n\left( C\cap A \right)+n\left( A\cap B\cap C \right)\].