Question

Let \( A = \{ n \in [100, 700] \cap \mathbb{N} : n \text{ is neither a multiple of 3 nor a multiple of 4} \} \).
Then the number of elements in \( A \) is:

• A. 300
• B. 280
• C. 310
• D. 290

Answer 1

The Correct Option is A

Step 1: Find the total number of elements in [100, 700]:

\[ \text{Total} = 700 - 100 + 1 = 601. \]

Step 2: Find the number of multiples of 3 in [100, 700]:

Multiples of 3: \( 102, 105, 108, \ldots, 699 \).
This is an arithmetic progression (AP) with: \[ a = 102, \, d = 3, \, \text{and } l = 699. \] The \(n\)-th term is: \[ T_n = a + (n - 1)d \implies 699 = 102 + (n - 1)3. \] Simplify: \[ 597 = 3(n - 1) \implies n = 200. \] Thus, \( n(3) = 200 \).

Step 3: Find the number of multiples of 4 in [100, 700]:

Multiples of 4: \( 100, 104, 108, \ldots, 700 \).
This is an AP with: \[ a = 100, \, d = 4, \, \text{and } l = 700. \] The \(n\)-th term is: \[ T_n = a + (n - 1)d \implies 700 = 100 + (n - 1)4. \] Simplify: \[ 600 = 4(n - 1) \implies n = 151. \] Thus, \( n(4) = 151 \).

Step 4: Find the number of multiples of both 3 and 4 (i.e., multiples of 12):

Multiples of 12: \( 108, 120, 132, \ldots, 696 \).
This is an AP with: \[ a = 108, \, d = 12, \, \text{and } l = 696. \] The \(n\)-th term is: \[ T_n = a + (n - 1)d \implies 696 = 108 + (n - 1)12. \] Simplify: \[ 588 = 12(n - 1) \implies n = 50. \] Thus, \( n(3 \cap 4) = 50 \).

Step 5: Use the inclusion-exclusion principle to find \( n(3 \cup 4) \):

\[ n(3 \cup 4) = n(3) + n(4) - n(3 \cap 4). \] Substitute values: \[ n(3 \cup 4) = 200 + 151 - 50 = 301. \]

Step 6: Find the number of elements in \(A\) (neither multiples of 3 nor 4):

\[ n(A) = \text{Total} - n(3 \cup 4). \] Substitute values: \[ n(A) = 601 - 301 = 300. \]