Answer: Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.
Step-by-step explanation:
Since we have given that
Integers between 10000 and 99999 = 99999-10000+1=90000
n( divisible by 3) = [tex]\dfrac{90000}{3}=30000[/tex]
n( divisible by 5) = [tex]\dfrac{90000}{5}=18000[/tex]
n( divisible by 7) = [tex]\dfrac{90000}{7}=12857.14[/tex]
n( divisible by 3 and 5) = n(3∩5)=[tex]\dfrac{90000}{15}=6000[/tex]
n( divisible by 5 and 7) = n(5∩7) = [tex]\dfrac{90000}{35}=2571.42[/tex]
n( divisible by 3 and 7) = n(3∩7) = [tex]\dfrac{90000}{21}=4285.71[/tex]
n( divisible by 3,5 and 7) = n(3∩5∩7) = [tex]\dfrac{90000}{105}=857.14[/tex]
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)
[tex]=30000+18000+12857.14-6000-2571.42-4258.71+857.14\\\\=48884.15[/tex]
Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.
