Answer: Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.
Explanation:
Since we have given that
Integers between 10000 and 99999 = 99999-10000+1=90000
n( divisible by 3) =
![(90000)/(3)=30000](https://img.qammunity.org/2020/formulas/mathematics/college/rvlj7mxi1pxbia9abtvq6wq290akotwit2.png)
n( divisible by 5) =
![(90000)/(5)=18000](https://img.qammunity.org/2020/formulas/mathematics/college/lyi94z050dy9g8tudql9d4wymjs1xqo0ny.png)
n( divisible by 7) =
![(90000)/(7)=12857.14](https://img.qammunity.org/2020/formulas/mathematics/college/mh2bxul90sh66mzua4qtzu2ytwk75u5sfy.png)
n( divisible by 3 and 5) = n(3∩5)=
![(90000)/(15)=6000](https://img.qammunity.org/2020/formulas/mathematics/college/is8eoawq71cxafrd6itqa2na1g2fbekc27.png)
n( divisible by 5 and 7) = n(5∩7) =
![(90000)/(35)=2571.42](https://img.qammunity.org/2020/formulas/mathematics/college/xbs3oinnluvjc5qy8ak0krr6c4jjjkstjw.png)
n( divisible by 3 and 7) = n(3∩7) =
![(90000)/(21)=4285.71](https://img.qammunity.org/2020/formulas/mathematics/college/lgq98v6x727k0in2ld3ffx4czmnknhxg0j.png)
n( divisible by 3,5 and 7) = n(3∩5∩7) =
![(90000)/(105)=857.14](https://img.qammunity.org/2020/formulas/mathematics/college/jlcm04play9y4ghc35vec2uz4i8kq5erxz.png)
As we know the formula,
n(3∪5∪7)=n(3)+n(5)+n(7)-n(3∩5)-n(5∩7)-n(3∩7)+n(3∩5∩7)
![=30000+18000+12857.14-6000-2571.42-4258.71+857.14\\\\=48884.15](https://img.qammunity.org/2020/formulas/mathematics/college/rscnujgsum7vrgr960af9ltorv2w2p1dea.png)
Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.