Divisibility Properties of P and Q

The question asks us to determine which statement is always true when a number P is divisible by 2 and another number Q is divisible by 3. Understanding the concept of divisibility is key to solving this problem involving properties of numbers.

Understanding Divisibility

When a number is divisible by another number, it means that the first number can be divided by the second number without leaving a remainder.

• If P is divisible by 2, it means P is a multiple of 2. We can write P as \( P = 2m \), where \( m \) is an integer. Examples of P could be 2, 4, 6, 8, etc.
• If Q is divisible by 3, it means Q is a multiple of 3. We can write Q as \( Q = 3n \), where \( n \) is an integer. Examples of Q could be 3, 6, 9, 12, etc.

Analyzing the Given Options

Let's examine each option based on the fact that \( P = 2m \) and \( Q = 3n \).

Option 1: P + Q is divisible by 6

\( P + Q = 2m + 3n \). Is this always divisible by 6? Let's test with examples:

• If \( P = 2 \) ( \( m=1 \) ) and \( Q = 3 \) ( \( n=1 \) ), then \( P + Q = 2 + 3 = 5 \). 5 is not divisible by 6.
• If \( P = 4 \) ( \( m=2 \) ) and \( Q = 3 \) ( \( n=1 \) ), then \( P + Q = 4 + 3 = 7 \). 7 is not divisible by 6.

Since we found examples where \( P + Q \) is not divisible by 6, this statement is not always true.

Option 2: P + Q is divisible by 5

\( P + Q = 2m + 3n \). Is this always divisible by 5? Let's test with examples:

• If \( P = 2 \) ( \( m=1 \) ) and \( Q = 3 \) ( \( n=1 \) ), then \( P + Q = 2 + 3 = 5 \). 5 is divisible by 5.
• If \( P = 4 \) ( \( m=2 \) ) and \( Q = 3 \) ( \( n=1 \) ), then \( P + Q = 4 + 3 = 7 \). 7 is not divisible by 5.

Since we found an example where \( P + Q \) is not divisible by 5, this statement is not always true.

Option 3: P × Q is divisible by 5

\( P \times Q = (2m) \times (3n) \). Is this always divisible by 5? \( P \times Q = 6mn \). Let's test with examples:

• If \( P = 2 \) ( \( m=1 \) ) and \( Q = 3 \) ( \( n=1 \) ), then \( P \times Q = 2 \times 3 = 6 \). 6 is not divisible by 5.
• If \( P = 4 \) ( \( m=2 \) ) and \( Q = 6 \) ( \( n=2 \) ), then \( P \times Q = 4 \times 6 = 24 \). 24 is not divisible by 5.

Since we found examples where \( P \times Q \) is not divisible by 5, this statement is not always true.

Option 4: P × Q is divisible by 6

\( P \times Q = (2m) \times (3n) \). Using the commutative and associative properties of multiplication, we can rearrange this as: \( P \times Q = (2 \times 3) \times (m \times n) \) \( P \times Q = 6 \times (mn) \)

Since \( m \) and \( n \) are integers, their product \( mn \) is also an integer. Therefore, \( P \times Q \) can be written as 6 multiplied by an integer (\( mn \)). By definition, this means \( P \times Q \) is always a multiple of 6, and thus, it is always divisible by 6.

This statement is always true.

Step-by-Step Solution for P times Q Divisibility

Here is a step-by-step breakdown for analyzing the product \( P \times Q \):

1. Identify the given information: P is divisible by 2, and Q is divisible by 3.
2. Write P in terms of its divisor: \( P = 2 \times m \) for some integer \( m \).
3. Write Q in terms of its divisor: \( Q = 3 \times n \) for some integer \( n \).
4. Consider the product \( P \times Q \): Substitute the expressions for P and Q. \( P \times Q = (2m) \times (3n) \).
5. Simplify the expression: \( P \times Q = 2 \times m \times 3 \times n = (2 \times 3) \times (m \times n) = 6 \times (mn) \).
6. Analyze the result: The product \( P \times Q \) is equal to 6 times the integer \( mn \).
7. Conclusion: Any number that can be expressed as 6 times an integer is divisible by 6. Therefore, \( P \times Q \) is always divisible by 6.

Conclusion on Divisibility

Based on the analysis of each option and the step-by-step derivation, the only statement that is always true is that \( P \times Q \) is divisible by 6, given that P is divisible by 2 and Q is divisible by 3.

Revision Table: Essential Divisibility Rules

Additional Information: Number Properties and Factors

Understanding factors and multiples is fundamental to divisibility. A factor of a number divides it evenly. A multiple of a number is the result of multiplying that number by an integer.

• Since P is divisible by 2, 2 is a factor of P, and P is a multiple of 2.
• Since Q is divisible by 3, 3 is a factor of Q, and Q is a multiple of 3.
• The product \( P \times Q \) includes both 2 (from P) and 3 (from Q) as factors.
• Since \( P \times Q \) has both 2 and 3 as factors, and 2 and 3 are prime numbers, their product (2 x 3 = 6) must also be a factor of \( P \times Q \).
• This confirms that \( P \times Q \) is divisible by 6.

This principle extends to other composite divisors. If a number is divisible by two coprime numbers (numbers with no common factors other than 1), it is also divisible by their product. In this case, 2 and 3 are coprime, so if a number is divisible by both 2 and 3, it's divisible by 2 x 3 = 6. P x Q is divisible by 2 (because P is) and P x Q is divisible by 3 (because Q is), so P x Q must be divisible by 6.