Evaluating Number Theory Statements on HCF and LCM

This solution analyzes two statements related to the properties of Highest Common Factor (HCF) and Least Common Multiple (LCM) in number theory. We are given specific conditions: \(a = bc\) and \(HCF(b, c) = 1\). The task is to determine which of the following statements are correct:

• Statement 1: If \(a = bc\) with \(HCF(b, c) = 1\), then \(HCF(c, bd) = HCF(c, d)\).
• Statement 2: If \(a = bc\) with \(HCF(b, c) = 1\), then \(LCM(a, d) = LCM(c, bd)\).

Analysis of Statement 1: Verifying HCF(c, bd) = HCF(c, d)

We start with the given conditions: \(a = bc\) and \(HCF(b, c) = 1\). The condition \(HCF(b, c) = 1\) signifies that \(b\) and \(c\) are coprime, meaning they share no common prime factors.

We need to check the equality \(HCF(c, bd) = HCF(c, d)\).

Let's use the properties of HCF. We can utilize a known property: If \(HCF(x, y) = 1\), then \(HCF(x, yz) = HCF(x, z)\).

In our case, we want to evaluate \(HCF(c, bd)\). We are given \(HCF(b, c) = 1\). This is equivalent to \(HCF(c, b) = 1\).

Let \(x=c\), \(y=b\), and \(z=d\). Applying the property, since \(HCF(c, b) = 1\), we get:

This matches Statement 1 exactly.

Therefore, Statement 1 is correct.

Analysis of Statement 2: Verifying LCM(a, d) = LCM(c, bd)

Again, we are given \(a = bc\) and \(HCF(b, c) = 1\). We need to verify if \(LCM(a, d) = LCM(c, bd)\) holds true.

First, substitute \(a = bc\) into the equation. We need to check if \(LCM(bc, d) = LCM(c, bd)\).

We can use the fundamental relationship between LCM and HCF: \(LCM(x, y) = \frac{|x \cdot y|}{HCF(x, y)}\).

Let's express the left and right sides using this formula:

• Left side: \(LCM(bc, d) = \frac{bc \cdot d}{HCF(bc, d)}\)
• Right side: \(LCM(c, bd) = \frac{c \cdot bd}{HCF(c, bd)}\)

From our analysis of Statement 1, we know \(HCF(c, bd) = HCF(c, d)\). Substituting this into the right side:

Now let's simplify the left side's denominator, \(HCF(bc, d)\). We use another property: If \(HCF(x, y) = 1\), then \(HCF(xy, z) = HCF(x, z) \cdot HCF(y, z)\).

Since we are given \(HCF(b, c) = 1\), we can apply this property with \(x=b\), \(y=c\), and \(z=d\):

Now substitute this back into the expression for the left side:

We compare the two sides:

For these two expressions to be equal, we would require:

This implies \(HCF(b, d) = 1\). However, the initial condition \(HCF(b, c) = 1\) does not guarantee that \(HCF(b, d)\) is also 1. \(b\) and \(d\) might share common factors.

Let's illustrate with a concrete counterexample:

• Choose \(b=2\), \(c=3\), and \(d=4\).
• Verify the conditions: \(a = bc = 2 \times 3 = 6\). \(HCF(b, c) = HCF(2, 3) = 1\). The conditions are met.
• Calculate \(LCM(a, d)\): \(LCM(6, 4)\). The prime factors are \(6 = 2 \times 3\) and \(4 = 2^2\). So, \(LCM(6, 4) = 2^2 \times 3 = 12\).
• Calculate \(LCM(c, bd)\): \(LCM(3, 2 \times 4) = LCM(3, 8)\). The prime factors are \(3 = 3\) and \(8 = 2^3\). So, \(LCM(3, 8) = 2^3 \times 3 = 24\).
• Compare the results: \(LCM(a, d) = 12\) and \(LCM(c, bd) = 24\). Since \(12 \neq 24\), the equality \(LCM(a, d) = LCM(c, bd)\) does not hold true in this case.

Therefore, Statement 2 is incorrect.

Final Conclusion

Summarizing our findings:

• Statement 1 (\(HCF(c, bd) = HCF(c, d)\)) is proven to be correct using properties of HCF.
• Statement 2 (\(LCM(a, d) = LCM(c, bd)\)) is found to be incorrect, as demonstrated by a counterexample.

Since only Statement 1 is correct, the correct option is 1 only.