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: Determine the total count of numbers in the range [100, 700].

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

Step 2: Calculate the number of multiples of 3 within [100, 700].

Multiples of 3: \( 102, 105, 108, \ldots, 699 \). This sequence forms an arithmetic progression (AP) with first term \( a = 102 \), common difference \( d = 3 \), and last term \( l = 699 \). Using the \(n\)-th term formula: \[ T_n = a + (n - 1)d \implies 699 = 102 + (n - 1)3. \] Solving for \(n\): \[ 597 = 3(n - 1) \implies n = 200. \] Therefore, the count of multiples of 3 is \( n(3) = 200 \).

Step 3: Calculate the number of multiples of 4 within [100, 700].

Multiples of 4: \( 100, 104, 108, \ldots, 700 \). This sequence forms an AP with first term \( a = 100 \), common difference \( d = 4 \), and last term \( l = 700 \). Using the \(n\)-th term formula: \[ T_n = a + (n - 1)d \implies 700 = 100 + (n - 1)4. \] Solving for \(n\): \[ 600 = 4(n - 1) \implies n = 151. \] Therefore, the count of multiples of 4 is \( n(4) = 151 \).

Step 4: Determine the count of numbers that are multiples of both 3 and 4 (i.e., multiples of 12) within [100, 700].

Multiples of 12: \( 108, 120, 132, \ldots, 696 \). This sequence forms an AP with first term \( a = 108 \), common difference \( d = 12 \), and last term \( l = 696 \). Using the \(n\)-th term formula: \[ T_n = a + (n - 1)d \implies 696 = 108 + (n - 1)12. \] Solving for \(n\): \[ 588 = 12(n - 1) \implies n = 50. \] Thus, the count of multiples of both 3 and 4 is \( n(3 \cap 4) = 50 \).

Step 5: Apply the inclusion-exclusion principle to find the count of numbers that are multiples of 3 or 4, denoted as \( n(3 \cup 4) \).

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

Step 6: Calculate the count of numbers in the set \(A\), which are neither multiples of 3 nor 4.

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