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$\frac{500}{3}+\frac{500}{5}+\frac{500}{7}-\frac{500}{15}-\frac{500}{35}-\frac{500} {21} +\frac{500}{105}$

Taking floor of each and every term

$66+100+71-33-14-23+4 = 271$

Option c) is correct

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Let A = number divisible by 3
B = numbers divisible by 5
C = number divisible by 7
We need to find “The number of integers between 1 and 500 that are divisible by 3 or 5 or 7" i.e.,|A∪B∪C|
We know,
|A∪B∪C|=|A|+|B|+C-|A∩B|-|A∩C|-|B∩C|+|A∩B|
|A|=number of integers divisible by 3
[500/3=166.6≈166=166]
|B|=100
[500/5=100]
|C|=71
[500/7=71.42]
|A∩B|=number of integers divisible by both 3 and 5 we need to compute with LCM (15)
i.e.,⌊500/15⌋≈33
|A∩B|=33
|A∩C|=500/LCM(3,7) 500/21=23.8≈28
|B∩C|=500/LCM(5,3) =500/35=14.48≈14
|A∩B∩C|=500/LCM(3,5,7) =500/163=4.76≈4
|A∪B∪C|=|A|+|B|+|C|-|A∩B|-|A∩C|-|B∩C|+|A∩B∩C|
=166+100+71-33-28-14+4
=271
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no of integer divisible by 3 in rangeof (1, 500) = floor( 500 / 3) = 166

no of integer divisible by 5 in rangeof(1, 500) = floor( 500 / 5) = 100

no of integer divisible by 7 in rageof(1, 500) = floor( 500 / 7) = 71

no of integer by 3 and 5 = floor{ 500 / L.C.M(3, 5)} = 33

no of integer by 5 and 7 = floor{ 500 / L.C.M(5, 7)} = 14

no of integer by 3 and 7 = floor{ 500 / L.C.M(3,7) } = 23

no of integer by 3 and 5 and 7  = floor{ 500 / L.C.M(3,5,7)} = 4

 

so, total no of integer divisible by 3 or 5 or 7 in rageof(1, 500) =

no of integer by 3 + no of integer by 5 + no of integer by 7 - { (no of integer by 3 and 5) -  (no of integer by 5 and 7) -  (no of integer by 3 and 7)  +  (no of integer by 3 and 5 and 7)

= 166 + 100 + 71 - 33 - 14 -23 + 4 = 271

 

So, "C is Correct"
Answer:

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