What is the number of natural numbers less than or equal to 1000 which are neither divisible by 10 nor 15 nor 25 NDA NA 2022 A 860 B 854 C 840 D 824?

The number of numbers divisible by $10$ is $[\frac{1000}{10}]=100$.

The number of numbers divisible by $15$ is $[\frac{1000}{15}]=66$.

The number of numbers divisible by $25$ is $[\frac{1000}{25}]=40$.

The number of numbers divisible by both $10$ and $15$ is $[\frac{1000}{30}]=33$.

The number of numbers divisible by both $10$ and $25$ is $[\frac{1000}{50}]=20$.

The number of numbers divisible by both $15$ and $25$ is $[\frac{1000}{75}]=13$.

The number of numbers divisible by $10$, $15$ and $25$ is $[\frac{1000}{150}]=6$.

Hence there are $100+66+40-33-20-13+6=146$ which are divisible by one of your numbers. Finally, there are $1000-146=854$ numbers which are not.

A little more insight:

The idea is that the first $3$ numbers $(100,66,50)$ represent how many integers in your range are divisible by the respective integers. However, some of them are counted twice, that's why we need to see how many of them appear in $2$ of the respective piles of numbers - for example, all numbers divisible by both $10$ and $15$ will appear in both the first and the second pile. Thus, after we find the number of such integers, we have to substract it. Finally, there are numbers that have appeared $3$ times initially and have been substracted $3$ times - hence we need to add them to the list. Those are precisely the numbers that are divisible by all of $10,15$ and $25$.

If you are more interested, have a look at the Inclusion-Exclusion Formula.

Concept:

• \({\rm{n}}\left( {{\rm{A\;U\;B\;U\;C}}} \right) = {\rm{\;n}}\left( {\rm{A}} \right) + {\rm{\;n}}\left( {\rm{B}} \right) + {\rm{\;n}}\left( {\rm{C}} \right) - {\rm{\;n}}\left( {{\rm{A}} \cap {\rm{B}}} \right) - {\rm{\;n}}\left( {{\rm{A}} \cap {\rm{C}}} \right) - {\rm{\;n}}\left( {{\rm{B}} \cap {\rm{C}}} \right) + {\rm{n}}\left( {{\rm{A}} \cap {\rm{B}} \cap {\rm{C}}} \right)\)

Calculation:

Let,

A = Number divisible by 10,

B = Number divisible by 15,

C = Number divisible by 25

\({\rm{n}}\left( {\rm{A}} \right) = \frac{{1000}}{{10}} = 100,{\rm{\;}}\)

\({\rm{n}}\left( {\rm{B}} \right) = \frac{{1000}}{{15}} = 66,{\rm{\;}}\)

\({\rm{n}}\left( {\rm{C}} \right) = \frac{{1000}}{{25}} = 40\)

Now, number divisible by A and B, i.e., n (A ∩ B)

For that we need to find out the least number which is divisible by both 10 and 15

And that is 30

\(\therefore {\rm{n}}\left( {{\rm{A}} \cap {\rm{B}}} \right) = \frac{{1000}}{{30}} = 33\)

Similarly,

\({\rm{n}}\left( {{\rm{A}} \cap {\rm{C}}} \right) = \frac{{1000}}{{50}} = 20\)                    (∵ The least number which is divisible by both 10 and 25 is 50)

Also,

\({\rm{n}}\left( {{\rm{B}} \cap {\rm{C}}} \right) = \frac{{1000}}{{75}} = 13\)                    (∵ The least number which is divisible by both 15 and 25 is 75)

And

The least number which is divisible by 10, 15 and 25 is 150

\({\rm{n}}\left( {{\rm{A}} \cap {\rm{B}} \cap {\rm{C}}} \right) = \frac{{1000}}{{150}} = 6\)

Number divisible by 10, 15 or 25

I.e., \({\rm{n}}\left( {{\rm{A\;U\;B\;U\;C}}} \right) = {\rm{\;n}}\left( {\rm{A}} \right) + {\rm{\;n}}\left( {\rm{B}} \right) + {\rm{\;n}}\left( {\rm{C}} \right) - {\rm{\;n}}\left( {{\rm{A}} \cap {\rm{B}}} \right) - {\rm{\;n}}\left( {{\rm{A}} \cap {\rm{C}}} \right) - {\rm{\;n}}\left( {{\rm{B}} \cap {\rm{C}}} \right) + {\rm{n}}\left( {{\rm{A}} \cap {\rm{B}} \cap {\rm{C}}} \right)\)

\(\therefore {\rm{n}}\left( {{\rm{A\;U\;B\;U\;C}}} \right) = 100 + 66 + 40 - 33 - 20 - 13 + 6\)

\( \Rightarrow {\rm{n}}\left( {{\rm{A\;U\;B\;U\;C}}} \right) = 146\)

Now, the number divisible neither by 10 nor 15, 25

= 1000 – 146

= 854

Hence, option (2) is correct.