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

Correct Option

The problem requires identifying the count of natural numbers from 1 to 100 that are not divisible by any of the numbers 2, 3, 5, 7, and 9.

First, simplify the set of divisors {2, 3, 5, 7, 9}. Since any number divisible by 9 is also divisible by 3, the condition "not divisible by 9" is automatically satisfied if the number is "not divisible by 3". Therefore, the problem reduces to finding numbers from 1 to 100 that are not divisible by any of the prime numbers: 2, 3, 5, or 7.

This can be solved using the Principle of Inclusion-Exclusion. Let N(d) denote the count of numbers from 1 to 100 divisible by d. The total number of integers is 100.

The count of numbers divisible by at least one of 2, 3, 5, or 7 is given by:

N(2∪3∪5∪7) = ΣN(p) - ΣN(p₁p₂) + ΣN(p₁p₂p₃) - N(p₁p₂p₃p₄)

• Numbers divisible by a single prime (p):
  • N(2) = ⌊100/2⌋ = 50
  • N(3) = ⌊100/3⌋ = 33
  • N(5) = ⌊100/5⌋ = 20
  • N(7) = ⌊100/7⌋ = 14
  • Sum (S₁) = 50 + 33 + 20 + 14 = 117
• Numbers divisible by the product of two distinct primes (p₁p₂), which is their Least Common Multiple (LCM):
  • N(2×3) = N(6) = ⌊100/6⌋ = 16
  • N(2×5) = N(10) = ⌊100/10⌋ = 10
  • N(2×7) = N(14) = ⌊100/14⌋ = 7
  • N(3×5) = N(15) = ⌊100/15⌋ = 6
  • N(3×7) = N(21) = ⌊100/21⌋ = 4
  • N(5×7) = N(35) = ⌊100/35⌋ = 2
  • Sum (S₂) = 16 + 10 + 7 + 6 + 4 + 2 = 45
• Numbers divisible by the product of three distinct primes (p₁p₂p₃):
  • N(2×3×5) = N(30) = ⌊100/30⌋ = 3
  • N(2×3×7) = N(42) = ⌊100/42⌋ = 2
  • N(2×5×7) = N(70) = ⌊100/70⌋ = 1
  • N(3×5×7) = N(105) = ⌊100/105⌋ = 0
  • Sum (S₃) = 3 + 2 + 1 + 0 = 6
• Numbers divisible by the product of four distinct primes (p₁p₂p₃p₄):
  • N(2×3×5×7) = N(210) = ⌊100/210⌋ = 0
  • Sum (S₄) = 0

Total numbers divisible by at least one of 2, 3, 5, or 7:

N(2∪3∪5∪7) = S₁ - S₂ + S₃ - S₄ = 117 - 45 + 6 - 0 = 72 + 6 = 78.

The number of integers from 1 to 100 that are not divisible by any of 2, 3, 5, or 7 is:

Total numbers - N(2∪3∪5∪7) = 100 - 78 = 22.

These 22 numbers include 1 and all prime numbers greater than 7 and less than or equal to 100 (i.e., 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97).

Incorrect Options

Options 1 (20), 2 (21), and 4 (23) are incorrect. These values would typically result from errors in calculation, misapplication of the Principle of Inclusion-Exclusion, or failure to correctly simplify the set of divisors (e.g., by not recognizing the redundancy of 9 when 3 is already considered). Any arithmetic inaccuracies during the summation and subtraction steps would lead to these incorrect counts.