Understanding the Problem: LCM and HCF

This question asks us to find two numbers based on their Least Common Multiple (LCM) and Highest Common Factor (HCF), with an additional condition that both numbers must be greater than 21. Once we find these numbers, we need to calculate their sum.

The key concepts involved are HCF, LCM, and the fundamental relationship between them for two numbers. The relationship states that the product of two numbers is equal to the product of their HCF and LCM.

Relating HCF, LCM, and the Numbers

Let the two unknown numbers be \(a\) and \(b\).

We are given:

• HCF of \(a\) and \(b\) = 21
• LCM of \(a\) and \(b\) = 3003
• Both \(a > 21\) and \(b > 21\)

Since the HCF of the two numbers is 21, we can express the numbers as multiples of their HCF. Let the numbers be \(a = 21x\) and \(b = 21y\), where \(x\) and \(y\) are co-prime integers. Co-prime means that the only positive integer that divides both \(x\) and \(y\) is 1. This is a crucial property when working with HCF.

Using the LCM-HCF Relationship to Find the Numbers

The fundamental relationship between two numbers, their HCF, and their LCM is:

Substitute the values we know into this equation:

Simplify the left side:

Now, let's calculate the product on the right side:

So, the equation becomes:

To find the value of the product \(xy\), divide 63063 by 441:

Let's perform the division:

Now we need to find pairs of co-prime integers \((x, y)\) whose product is 143.

Let's find the factors of 143. We can check small prime numbers. It's not divisible by 2, 3, 5, 7. Let's try 11. \(143 \div 11 = 13\). Since 11 and 13 are prime numbers, their only factors are 1, 11, 13, 143. The pairs of factors whose product is 143 are \((1, 143)\) and \((11, 13)\). Both of these pairs consist of co-prime numbers.

Checking the Condition: Numbers Greater Than 21

We have two possible pairs for \((x, y)\):

1. \((x, y) = (1, 143)\)
2. \((x, y) = (11, 13)\)

Let's find the corresponding numbers \(a = 21x\) and \(b = 21y\) for each pair and check if they satisfy the condition that both numbers are greater than 21.

Case 1: \((x, y) = (1, 143)\)

• Number 1: \(a = 21 \times x = 21 \times 1 = 21\)
• Number 2: \(b = 21 \times y = 21 \times 143 = 3003\)

In this case, one number is 21 and the other is 3003. The condition is that both numbers must be greater than 21. Since 21 is not greater than 21, this pair of numbers does not satisfy the condition.

Case 2: \((x, y) = (11, 13)\)

• Number 1: \(a = 21 \times x = 21 \times 11\)
• Number 2: \(b = 21 \times y = 21 \times 13\)

Let's calculate these values:

• \(21 \times 11 = 231\)
• \(21 \times 13 = 273\)

The two numbers are 231 and 273. Let's check the condition: Is 231 > 21? Yes. Is 273 > 21? Yes. Both numbers satisfy the condition.

Therefore, the two numbers are 231 and 273.

Calculating the Sum of the Numbers

The question asks for the sum of these two numbers.

Sum = \(a + b = 231 + 273\)

Let's add them:

The sum of the two numbers is 504.

Summary of Steps

1. Represent the two numbers as \(21x\) and \(21y\), where \(x\) and \(y\) are co-prime, since their HCF is 21.
2. Use the relation \(a \times b = \text{LCM} \times \text{HCF}\) to find the product \(xy\).
3. Find co-prime pairs \((x, y)\) whose product is \(xy\).
4. For each pair, calculate the two numbers \(21x\) and \(21y\).
5. Check which pair satisfies the condition that both numbers are greater than 21.
6. Calculate the sum of the numbers from the valid pair.

The numbers are 231 and 273.

Their sum is \(231 + 273 = 504\).

Revision Table: LCM and HCF Concepts

Additional Information: Co-prime Numbers

Two integers \(x\) and \(y\) are said to be co-prime or relatively prime if their only positive common divisor is 1. This means their HCF is 1. When two numbers \(a\) and \(b\) have an HCF of \(h\), they can be written as \(a = hx\) and \(b = hy\), where \(x\) and \(y\) must be co-prime. This property is essential for solving problems like this one.

For example, if the numbers are 12 and 18, their HCF is 6. We can write \(12 = 6 \times 2\) and \(18 = 6 \times 3\). Here, \(x=2\) and \(y=3\). 2 and 3 are co-prime.