A = {n ∈ [100,700] ∩ N: n is neither a multiple of 3 nor of 4}. Find no. of elements in A

To find the number of elements in set \( A \), where \( A = \{ n \in [100, 700] \cap \mathbb{N} : n \text{ is neither a multiple of 3 nor of 4} \} \), follow these steps:

### Step 1: Determine the total number of natural numbers between 100 and 700.
The natural numbers between 100 and 700 (inclusive) are:
\[
700 - 100 + 1 = 601
\]
Thus, there are 601 natural numbers in this range.

### Step 2: Find the number of multiples of 3 in the range.
The formula to find multiples of a number \( x \) between \( a \) and \( b \) is:
\[
\left\lfloor \frac{b}{x} \right\rfloor - \left\lfloor \frac{a-1}{x} \right\rfloor
\]
For multiples of 3 between 100 and 700:
\[
\left\lfloor \frac{700}{3} \right\rfloor = 233 \quad \text{and} \quad \left\lfloor \frac{99}{3} \right\rfloor = 33
\]
Thus, the number of multiples of 3 is:
\[
233 - 33 = 200
\]

### Step 3: Find the number of multiples of 4 in the range.
Similarly, for multiples of 4 between 100 and 700:
\[
\left\lfloor \frac{700}{4} \right\rfloor = 175 \quad \text{and} \quad \left\lfloor \frac{99}{4} \right\rfloor = 24
\]
Thus, the number of multiples of 4 is:
\[
175 - 24 = 151
\]

### Step 4: Find the number of multiples of 12 in the range.
Since 12 is the least common multiple (LCM) of 3 and 4, we apply the same formula for multiples of 12:
\[
\left\lfloor \frac{700}{12} \right\rfloor = 58 \quad \text{and} \quad \left\lfloor \frac{99}{12} \right\rfloor = 8
\]
Thus, the number of multiples of 12 is:
\[
58 - 8 = 50
\]

### Step 5: Apply the inclusion-exclusion principle.
Using inclusion-exclusion, the number of numbers that are either multiples of 3 or multiples of 4 is:
\[
( \text{Multiples of 3} ) + ( \text{Multiples of 4} ) - ( \text{Multiples of 12} )
\]
\[
200 + 151 - 50 = 301
\]

### Step 6: Find the number of elements in set \( A \).
The elements in \( A \) are the numbers that are neither multiples of 3 nor multiples of 4, which is the complement of the set we just found. Therefore, the number of elements in \( A \) is:
\[
601 - 301 = 300
\]

Thus, the number of elements in set \( A \) is \( \boxed{300} \).

Would you like any further details or have any questions?

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Here are 5 related questions to expand your understanding:
1. How would the solution change if \( A \) included numbers divisible by 4 but not 3?
2. How many numbers between 100 and 700 are multiples of both 5 and 6?
3. What is the number of integers between 200 and 900 that are not divisible by 7?
4. How would you modify the method to find numbers divisible by 5 but not by 10 between two numbers?
5. How does the inclusion-exclusion principle work in general for more than two sets?

**Tip:** The inclusion-exclusion principle is a powerful tool for finding the union of sets, especially when there are overlapping elements.

A = {n ∈ [100,700] ∩ N: n is neither a multiple of 3 nor of 4}. Find no. of elements in A.

Learn how to determine the number of natural numbers between 100 and 700 that are neither multiples of 3 nor multiples of 4. This problem uses the inclusion-exclusion principle and formulas to find multiples of numbers within a range, suitable for high school students.

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