Finding Pairs of Numbers with Given Product and HCF

The problem asks us to find the number of possible pairs of two numbers whose product is 1500 and whose Highest Common Factor (HCF) is 10.

Let the two numbers be \(a\) and \(b\). We are given:

• Product of the numbers: \(a \times b = 1500\)
• HCF of the numbers: HCF\((a, b) = 10\)

A key property relating HCF and numbers is that if the HCF of two numbers \(a\) and \(b\) is \(h\), then the numbers can be written as \(a = hx\) and \(b = hy\), where \(x\) and \(y\) are coprime integers. This means the HCF of \(x\) and \(y\) is 1 (HCF\((x, y) = 1\)).

Using this property, we can write our numbers as:

• \(a = 10x\)
• \(b = 10y\)

where \(x\) and \(y\) are coprime integers.

Now, let's use the given product of the numbers:

Divide both sides by 100 to find the value of \(xy\):

So, we need to find pairs of integers \((x, y)\) such that their product is 15 and they are coprime (HCF\((x, y) = 1\)).

Let's list the pairs of positive integers whose product is 15:

• (1, 15)
• (3, 5)
• (5, 3)
• (15, 1)

Now, let's check the HCF for each pair \((x, y)\) to see which ones are coprime:

• For (1, 15): HCF(1, 15) = 1. This pair is coprime.
• For (3, 5): HCF(3, 5) = 1. This pair is coprime.
• For (5, 3): HCF(5, 3) = 1. This pair is coprime.
• For (15, 1): HCF(15, 1) = 1. This pair is coprime.

The pairs \((x, y)\) that satisfy both conditions (\(xy=15\) and HCF\((x, y) = 1\)) are (1, 15), (3, 5), (5, 3), and (15, 1).

These pairs of \((x, y)\) correspond to the pairs of numbers \((a, b)\) using \(a = 10x\) and \(b = 10y\):

• If \((x, y) = (1, 15)\), then \(a = 10 \times 1 = 10\) and \(b = 10 \times 15 = 150\). The pair of numbers is (10, 150).
• If \((x, y) = (3, 5)\), then \(a = 10 \times 3 = 30\) and \(b = 10 \times 5 = 50\). The pair of numbers is (30, 50).
• If \((x, y) = (5, 3)\), then \(a = 10 \times 5 = 50\) and \(b = 10 \times 3 = 30\). The pair of numbers is (50, 30).
• If \((x, y) = (15, 1)\), then \(a = 10 \times 15 = 150\) and \(b = 10 \times 1 = 10\). The pair of numbers is (150, 10).

The question asks for the number of possible *pairs* of numbers. The pairs (10, 150) and (150, 10) represent the same pair of numbers {10, 150}. Similarly, (30, 50) and (50, 30) represent the same pair of numbers {30, 50}.

So, the unique pairs of numbers satisfying the conditions are {10, 150} and {30, 50}.

Let's verify the conditions for these pairs:

• Pair {10, 150}: Product \(10 \times 150 = 1500\). HCF(10, 150) = 10. (10 = \(2 \times 5\), 150 = \(10 \times 15 = 2 \times 5 \times 3 \times 5\). Common factors are 2 and 5. HCF = \(2 \times 5 = 10\)). This pair is valid.
• Pair {30, 50}: Product \(30 \times 50 = 1500\). HCF(30, 50) = 10. (30 = \(2 \times 3 \times 5\), 50 = \(2 \times 5 \times 5\). Common factors are 2 and 5. HCF = \(2 \times 5 = 10\)). This pair is valid.

There are exactly 2 such possible pairs of numbers.

From the table, we see two unique pairs of numbers: {10, 150} and {30, 50}.

Revision Table: Understanding Product, HCF, and Coprime Numbers

This problem utilizes the relationship between the product of two numbers, their HCF, and the concept of coprime numbers. Here's a quick recap:

• HCF (Highest Common Factor): The largest positive integer that divides two or more integers without leaving a remainder.
• Coprime Numbers: Two integers are coprime (or relatively prime) if their only positive common divisor is 1. HCF of coprime numbers is always 1.
• Relationship: If \(a\) and \(b\) are two numbers and their HCF is \(h\), then \(a = hx\) and \(b = hy\), where \(x\) and \(y\) are coprime integers (HCF\((x, y) = 1\)). The product \(a \times b = hx \times hy = h^2 xy\). This gives a useful formula: \(a \times b = \text{HCF}(a, b)^2 \times xy\). In this problem, \(1500 = 10^2 \times xy\), so \(1500 = 100xy\), leading to \(xy = 15\).

Additional Information: Factorization and Number Pairs

Finding pairs of numbers with specific properties often involves factorizing numbers and understanding the constraints like HCF. In this case, we factorized 15 to find pairs of coprime factors. Each pair of coprime factors \((x, y)\) of \(\frac{\text{Product}}{\text{HCF}^2}\) corresponds to a unique pair of numbers \((10x, 10y)\) with the given HCF and product.

For a general case, if the product of two numbers is \(P\) and their HCF is \(h\), then the numbers are \(hx\) and \(hy\) where HCF\((x, y) = 1\) and \(xy = \frac{P}{h^2}\). The number of such pairs is equal to the number of pairs of coprime factors of \(\frac{P}{h^2}\).

In our problem, \(\frac{P}{h^2} = \frac{1500}{10^2} = \frac{1500}{100} = 15\). The coprime factor pairs of 15 are (1, 15) and (3, 5). Each pair corresponds to a unique set of numbers {10, 150} and {30, 50}.

The number of such possible pairs is 2.