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

Answer: (A)

300

Answer 2

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: $n(A) = 601 - 301 = 300.$