GATE | GATE-CS-2017 (Set 1) | Question 58

The number of integers between 1 and 500 (both inclusive) that are divisible by 3 or 5 or 7 is ______.

Note: This questions appeared as Numerical Answer Type.
(A) 269
(B) 270
(C) 271
(D) 272


Answer: (C)

Explanation: The general formula for the union of 3 sets is:
(A union B union C) = A + B + C – (A intersect B) – (A intersect C) – (B intersect C) + (A intersect B intersect C).

Assuming,
A = 3, B = 5, C = 7
= 500/3 + 500/5 + 500/7 – 500/3*5 – 500/5*7 – 500/7*3 + 500/105
= 271
Therefore, option C is correct.



Alternate Solution:

Number of integers divisible by 3 or 5 or 7
= n (3 V 5 V 7) = n (3) + n(5) + n (7) – n (3 \wedge 5) – n (5 \wedge 7) -n (3 \wedge 7) + n (3 \wedge 5 \wedge 7)
= floor(500 1 3 )+ floor (50015) + floor(50017) – floor (500115) – floor (500/35) -floor(500121) + floor (500/105)
= 166 + 100 + 71 -33-14-23+4 = 271

This solution is contributed by Sumouli Chaudhary.



Alternate solution

Let a = number divisible by 3
b = number divisible by 5
c = number divisible by 7

n(a) = 166
n(b) = 100
n(c) = 71

n(a∩b) = number divisible by 15 = 33
n(b∩c) = number divisible by 35 = 14
n(a∩c) = number divisible by 21 = 23
n(a∩b∩c) = number divisible by 105 = 4
n(a∪b∪c) = n(a) + n(b) + n(c) – n(a∩b) – n(b∩c) – n(a∩c) + n(a∩b∩c) = 166 + 100 + 71 – 33 – 14 – 23 + 4 = 271



This solution is contributed by parul sharma

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