The sum of LCM and HCF of two numbers is 1484 and the difference between LCM and HCF is. 1428. If one of the numbers is 112, then what is the other number?
364
Understanding the Problem: Finding the Other Number
The question gives us information about the Lowest Common Multiple (LCM) and Highest Common Factor (HCF) of two numbers. We know their sum and their difference. We are also given one of the two numbers and need to find the other number.
We will use the following steps to solve this LCM and HCF problem:
- Find the values of LCM and HCF using the given sum and difference.
- Use the fundamental property relating two numbers, their LCM, and their HCF.
- Calculate the other number using this property.
Calculating LCM and HCF from Sum and Difference
Let the LCM of the two numbers be \(L\) and the HCF be \(H\).
According to the problem statement, we have two equations:
- Sum of LCM and HCF: \(L + H = 1484\) (Equation 1)
- Difference between LCM and HCF: \(L - H = 1428\) (Equation 2)
We can solve these two linear equations simultaneously to find \(L\) and \(H\).
Add Equation 1 and Equation 2:
\[ (L + H) + (L - H) = 1484 + 1428 \] \[ L + H + L - H = 2912 \] \[ 2L = 2912 \] \[ L = \frac{2912}{2} \] \[ L = 1456 \]
So, the LCM is 1456.
Now substitute the value of \(L\) into Equation 1:
\[ 1456 + H = 1484 \] \[ H = 1484 - 1456 \] \[ H = 28 \]
So, the HCF is 28.
Using the Property: Product of Numbers = LCM × HCF
A key property for any two positive integers is that the product of the two numbers is equal to the product of their LCM and HCF.
Let the two numbers be \(n_1\) and \(n_2\).
The property states: \(n_1 \times n_2 = \text{LCM} \times \text{HCF}\)
We are given that one of the numbers is 112. Let \(n_1 = 112\). We found LCM = 1456 and HCF = 28.
Substitute these values into the property equation:
\[ 112 \times n_2 = 1456 \times 28 \]
Now, we can solve for the other number, \(n_2\):
\[ n_2 = \frac{1456 \times 28}{112} \]
We can simplify this calculation. Notice that \(112 = 4 \times 28\). So, we can rewrite the expression as:
\[ n_2 = \frac{1456 \times 28}{4 \times 28} \]
Cancel out the 28 from the numerator and denominator:
\[ n_2 = \frac{1456}{4} \]
Now, perform the division:
\[ n_2 = 364 \]
Therefore, the other number is 364.
Verification
Let's check if the numbers 112 and 364 satisfy the conditions.
- HCF(112, 364): \(112 = 4 \times 28\), \(364 = 13 \times 28\). The HCF is 28.
- LCM(112, 364): Using the property \(n_1 \times n_2 = \text{LCM} \times \text{HCF}\), LCM = \(\frac{112 \times 364}{28} = 4 \times 364 = 1456\).
Now, check the sum and difference:
- Sum of LCM and HCF = \(1456 + 28 = 1484\). This matches the given sum.
- Difference between LCM and HCF = \(1456 - 28 = 1428\). This matches the given difference.
The calculations are consistent with the problem statement.
Summary of Calculation
| Description | Calculation | Result |
|---|---|---|
| Given Sum (L+H) | 1484 | |
| Given Difference (L-H) | 1428 | |
| \(2L = (L+H) + (L-H)\) | \(1484 + 1428\) | 2912 |
| \(L = 2912 / 2\) | 1456 | |
| \(H = (L+H) - L\) | \(1484 - 1456\) | 28 |
| One Number (\(n_1\)) | 112 | |
| LCM \(\times\) HCF | \(1456 \times 28\) | 40768 |
| \(n_2 = (\text{LCM} \times \text{HCF}) / n_1\) | \(40768 / 112\) | 364 |
Revision Table: Key Concepts in LCM and HCF
| Concept | Definition | Property |
|---|---|---|
| HCF (Highest Common Factor) | The largest positive integer that divides two or more integers without leaving a remainder. Also known as GCD (Greatest Common Divisor). | HCF of two numbers must divide their difference. |
| LCM (Lowest Common Multiple) | The smallest positive integer that is a multiple of two or more integers. | LCM of two numbers is divisible by each of the numbers. |
| Relationship between HCF, LCM, and Two Numbers | For any two positive integers \(n_1\) and \(n_2\). | \(n_1 \times n_2 = \text{HCF}(n_1, n_2) \times \text{LCM}(n_1, n_2)\) |
Additional Information: Solving Systems of Equations and Number Properties
This problem required solving a system of two linear equations with two variables (\(L\) and \(H\)). The method used, adding the two equations, is a common technique to eliminate one variable and solve for the other. Once one variable is found, it is substituted back into one of the original equations to find the second variable.
The property \(n_1 \times n_2 = \text{HCF} \times \text{LCM}\) is fundamental when dealing with LCM and HCF of two numbers. It is derived from the prime factorization of the numbers. If the prime factorizations of \(n_1\) and \(n_2\) are known, their HCF involves the minimum power of each common prime factor, while their LCM involves the maximum power of each prime factor present in either number. The product \(n_1 \times n_2\) combines all prime factors with powers that sum up the powers in \(n_1\) and \(n_2\), which is exactly what happens when you multiply the HCF and LCM.
Understanding these concepts and properties is crucial for solving problems involving LCM and HCF in competitive exams and quantitative aptitude tests.
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