Q. Consider the first 100 natural numbers. How many of them are not divisible by any one of 2, 3, 5, 7 and 9?

Explanation : 

We want the count of integers from 1 to 100 that are not divisible by any of 2, 3, 5, 7, or 9. Since any multiple of 9 is already a multiple of 3, excluding multiples of 3 covers multiples of 9 as well. So we only need to exclude multiples of 2, 3, 5, and 7.

Let
A₂ = {multiples of 2},
A₃ = {multiples of 3},
A₅ = {multiples of 5},
A₇ = {multiples of 7}.

We apply the Inclusion–Exclusion Principle to find how many are divisible by at least one of these, then subtract from 100.

1. Count of multiples of each single divisor
• |A₂| = ⌊100/2⌋ = 50
• |A₃| = ⌊100/3⌋ = 33
• |A₅| = ⌊100/5⌋ = 20
• |A₇| = ⌊100/7⌋ = 14
Sum₁ = 50 + 33 + 20 + 14 = 117

2. Subtract multiples counted twice (pairwise intersections)
• |A₂∩A₃| = ⌊100/lcm(2,3)=6⌋ = 16
• |A₂∩A₅| = ⌊100/10⌋ = 10
• |A₂∩A₇| = ⌊100/14⌋ = 7
• |A₃∩A₅| = ⌊100/15⌋ = 6
• |A₃∩A₇| = ⌊100/21⌋ = 4
• |A₅∩A₇| = ⌊100/35⌋ = 2
Sum₂ = 16 + 10 + 7 + 6 + 4 + 2 = 45

3. Add back multiples we subtracted too many times (triple intersections)
• |A₂∩A₃∩A₅| = ⌊100/30⌋ = 3
• |A₂∩A₃∩A₇| = ⌊100/42⌋ = 2
• |A₂∩A₅∩A₇| = ⌊100/70⌋ = 1
• |A₃∩A₅∩A₇| = ⌊100/105⌋ = 0
Sum₃ = 3 + 2 + 1 + 0 = 6

Note on Step 3:
After Step 1, a number divisible by, say, 2, 3, and 5 is counted three times.
In Step 2 we subtract it once for each pair (2–3, 2–5, 3–5), so we subtract it three times—bringing its net count to zero.
But in the union it should appear once, so we “add back” each triple‐intersection exactly once.

4. Four-way intersection
lcm(2,3,5,7)=210 > 100 ⇒ 0

By Inclusion–Exclusion, the total divisible by at least one of 2, 3, 5, 7 is
117 (singles)
– 45 (pairs)
+ 6 (triples)
– 0 (quadruple)
= 78.

Therefore the count not divisible by any of 2, 3, 5, 7, 9 is
100 – 78 = 22.

Final Answer: 22.