Understanding the Problem: Finding the Union of Two Sets

The question asks us to find the number of elements in the union of two sets, set A and set B. Set A contains the first 250 natural numbers that are multiples of 3. Set B contains the first 200 even natural numbers. To find the number of elements in the union of two sets ($A \cup B$), we use the Principle of Inclusion-Exclusion formula:

$$|A \cup B| = |A| + |B| - |A \cap B|$$

Here, $|A|$ is the number of elements in set A, $|B|$ is the number of elements in set B, and $|A \cap B|$ is the number of elements in the intersection of set A and set B.

Calculating the Number of Elements in Set A ($|A|$)

Set A consists of the first 250 natural numbers that are multiples of 3. These numbers are $3 \times 1, 3 \times 2, 3 \times 3, \dots$, up to the 250th multiple of 3. The 250th multiple of 3 is $3 \times 250 = 750$.

• Set A = $\{3, 6, 9, \dots, 750\}$
• The number of elements in Set A is exactly the count of the first 250 multiples of 3.
• Therefore, $|A| = 250$.

Calculating the Number of Elements in Set B ($|B|$)

Set B consists of the first 200 even natural numbers. Even natural numbers are multiples of 2. These numbers are $2 \times 1, 2 \times 2, 2 \times 3, \dots$, up to the 200th even number. The 200th even number is $2 \times 200 = 400$.

• Set B = $\{2, 4, 6, \dots, 400\}$
• The number of elements in Set B is exactly the count of the first 200 even natural numbers.
• Therefore, $|B| = 200$.

Calculating the Number of Elements in the Intersection ($|A \cap B|$)

The intersection of set A and set B, $A \cap B$, contains the elements that are present in both sets. An element in $A \cap B$ must be:

1. A multiple of 3 (because it's in set A).
2. An even number (a multiple of 2, because it's in set B).

A number that is both a multiple of 3 and a multiple of 2 is a multiple of the least common multiple (LCM) of 3 and 2. The LCM(3, 2) is 6. So, the elements in $A \cap B$ are multiples of 6.

Now we need to find how many such multiples of 6 are present in both sets A and B. The elements of set A are $\{3, 6, 9, \dots, 750\}$. The elements of set B are $\{2, 4, 6, \dots, 400\}$.

An element $x$ is in $A \cap B$ if it is a multiple of 6, and it is one of the first 250 multiples of 3, AND one of the first 200 even numbers.

Let $x = 6m$ for some natural number $m$.

• Since $x \in A$, $x$ must be of the form $3k$ where $1 \le k \le 250$. So, $6m = 3k \implies k = 2m$. For $1 \le k \le 250$, we have $1 \le 2m \le 250$, which means $0.5 \le m \le 125$. Since $m$ is a natural number, $1 \le m \le 125$.
• Since $x \in B$, $x$ must be of the form $2j$ where $1 \le j \le 200$. So, $6m = 2j \implies j = 3m$. For $1 \le j \le 200$, we have $1 \le 3m \le 200$, which means $1/3 \le m \le 200/3 \approx 66.67$. Since $m$ is a natural number, $1 \le m \le 66$.

For $x$ to be in both sets, $m$ must satisfy both conditions: $1 \le m \le 125$ and $1 \le m \le 66$. The common range for $m$ is $1 \le m \le 66$.

So, the elements in $A \cap B$ are the first 66 multiples of 6, which are $6 \times 1, 6 \times 2, \dots, 6 \times 66$. The largest element in $A \cap B$ is $6 \times 66 = 396$. Note that 396 is a multiple of 3 ($396 = 3 \times 132$, and $132 \le 250$) and it is an even number ($396 = 2 \times 198$, and $198 \le 200$).

• The number of elements in $A \cap B$ is the count of possible values for $m$.
• Therefore, $|A \cap B| = 66$.

Calculating the Number of Elements in the Union ($|A \cup B|$)

Now we use the Inclusion-Exclusion Principle:

$$|A \cup B| = |A| + |B| - |A \cap B|$$ $$|A \cup B| = 250 + 200 - 66$$ $$|A \cup B| = 450 - 66$$ $$|A \cup B| = 384$$

The number of elements in $A \cup B$ is 384.

Conclusion on Set Union Size

By applying the formula and calculating the size of set A, set B, and their intersection, we found that the total number of elements in the union of set A and set B is 384.

Revision Table: Key Set Theory Concepts

Additional Information on Counting Set Elements

When dealing with counting elements in sets, especially finite sets derived from number properties like multiples or even/odd numbers within a certain range or count, it's crucial to correctly identify the elements of each set and their intersection. The interpretation of phrases like "first N numbers that are..." is key. In this case, "first 250 natural numbers that are multiple of 3" means the set $\{3 \times 1, 3 \times 2, \dots, 3 \times 250\}$. Similarly, "first 200 even natural numbers" means $\{2 \times 1, 2 \times 2, \dots, 2 \times 200\}$.

Finding the intersection often involves finding numbers that satisfy multiple divisibility conditions (like being a multiple of both 3 and 2, meaning a multiple of LCM(3,2)). Then, you need to find how many such numbers exist within the constraints imposed by the definitions of the original sets A and B. The elements in the intersection must belong to both sets A and B. This translates to finding multiples of 6 that are of the form $3k$ with $k \le 250$ AND of the form $2j$ with $j \le 200$. The smallest value constraint on the index ($m$ in $6m$) determines the size of the intersection.