Question

If \(x\) and \(y\) are two positive integers, and \(m\) is the HCF of \(x\) and \(y\) such that \(mxy = 1080\) and \(3<m<12\), then how many possible ordered pairs of \(x\) and \(y\) exist?

• A. 9
• B. 2
• C. 6
• D. 18

Hint:

When dealing with HCF/LCM problems involving products, always substitute \(x = h \cdot a\) and \(y = h \cdot b\) where \(h\) is HCF and \(\text{gcd}(a, b) = 1\). This isolates the common factor and simplifies the problem.

Answer 1

The Correct Option is B

Step 1: Formulate the Equation:
Since \(m\) is the HCF of \(x\) and \(y\), we can write: \[ x = ma \quad \text{and} \quad y = mb \] where \(a\) and \(b\) are coprime positive integers (i.e., \(\text{HCF}(a, b) = 1\)). Substitute these into the given equation \(mxy = 1080\): \[ m(ma)(mb) = 1080 \] \[ m^3 ab = 1080 \] Step 2: Determine Possible Values of \(m\):
From \(m^3 ab = 1080\), \(m^3\) must be a perfect cube that divides 1080. Prime factorization of 1080: \[ 1080 = 108 \times 10 = 27 \times 4 \times 2 \times 5 = 3^3 \times 2^3 \times 5 \] The perfect cube factors of 1080 are \(1^3 = 1\), \(2^3 = 8\), \(3^3 = 27\), and \((2 \times 3)^3 = 6^3 = 216\). So, possible values for \(m\) are \(1, 2, 3, 6\). Step 3: Apply the Constraint:
We are given \(3<m<12\). From the set \(\{1, 2, 3, 6\}\), only \(m = 6\) satisfies the condition. Step 4: Find Pairs of \((a, b)\):
Substitute \(m = 6\) back into the equation: \[ 6^3 ab = 1080 \] \[ 216 ab = 1080 \] \[ ab = \frac{1080}{216} = 5 \] Since \(a\) and \(b\) must be coprime, the pairs \((a, b)\) that multiply to 5 are: 1. \((1, 5)\) 2. \((5, 1)\) (Note: 1 and 5 are coprime). Step 5: Find Ordered Pairs \((x, y)\):
Using \(x = 6a\) and \(y = 6b\): 1. For \((1, 5)\): \(x = 6, y = 30\). Pair: \((6, 30)\). 2. For \((5, 1)\): \(x = 30, y = 6\). Pair: \((30, 6)\). Total number of ordered pairs is 2.