Let A={n ∈[100,700] ∩ N: n is neither a multiple of 3 nor a multiple of 4}. Then the number of elements in A is

Solution

Maths sir · 3 steps

Count the numbers you do not want, the multiples of 3 or 4, then take them away from the total.

1. Total count. The natural numbers from 100 to 700, both included, number 700 - 100 + 1 = 601.
2. Count by subtraction. Multiples of k up to 700, minus multiples of k up to 99, gives the multiples of k in the range. A number divisible by both 3 and 4 is a multiple of 12.

   Multiples of	Up to 700	Up to 99	In [100, 700]
   3	⌊ 700/3 ⌋ = 233	⌊ 99/3 ⌋ = 33	200
   4	⌊ 700/4 ⌋ = 175	⌊ 99/4 ⌋ = 24	151
   12	⌊ 700/12 ⌋ = 58	⌊ 99/12 ⌋ = 8	50

   n(3 ∪ 4) = 200 + 151 - 50 = 301

3. So. Neither a multiple of 3 nor of 4: 601 - 301 = 300. The answer is (a).

The idea to keep. "Neither A nor B" means total minus n(A ∪ B), and n(A ∪ B) = n(A) + n(B) - n(A ∩ B). Remember that the overlap of multiples of 3 and 4 is the multiples of 12.

Answer from the NTA final answer key. Solution by PrepWiser.