Consider the following sentences: 1. If a = bc with HCF(b, c) = 1, then HCF(c, bd) = HCF(c, d). 2. If a = bc with HCF(b, c) = 1, then LCM(a, d) = LCM(c, bd).
1 only
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:
\[ HCF(c, bd) = HCF(c, d) \]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:
\[ LCM(c, bd) = \frac{cbd}{HCF(c, d)} \]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 \(z=d\):
\[ HCF(bc, d) = HCF(b, d) \cdot HCF(c, d) \]Now substitute this back into the expression for the left side:
\[ LCM(bc, d) = \frac{bcd}{HCF(b, d) \cdot HCF(c, d)} \]We compare the two sides:
\[ \frac{bcd}{HCF(b, d) \cdot HCF(c, d)} \quad \text{vs} \quad \frac{cbd}{HCF(c, d)} \]For these two expressions to be equal, we would require:
\[ \frac{1}{HCF(b, d) \cdot HCF(c, d)} = \frac{1}{HCF(c, d)} \]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.
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Important Questions from LCM and HCF
The HCF and LCM of two numbers are 12 and 72, respectively. If the ratio of the two numbers is 2 ∶ 3, then the larger of the two numbers is:
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Joseph visits the club on every 5 th day, Harsh visits on every 24 th day, while Sumit visits on every 9 th day. If all three of them met at the club on a Sunday, then on which day will all three of them meet again?
What is the least number which when divided by 12,20 and 24 leaves in each case a remainder of 8?
Which of the following is a pair of co-primes?
