Question

There are two numbers which are greater than 21 and their LCM and HCF are 3003 and 21 respectively. What is the sum of these numbers?

• A. 504
• B. 508
• C. 514
• D. 528

Answer 1

The Correct Option is A

To find the sum of the two numbers whose LCM is 3003 and HCF is 21, we can use the relationship between the LCM, HCF, and the numbers themselves. The formula that relates these quantities is:

\(LCM \times HCF = \text{Product of the Numbers}\)

Given: 

• \(LCM = 3003\)
• \(HCF = 21\)

Let the two numbers be \(a\) and \(b\). According to the formula:

\(3003 \times 21 = a \times b\)

Calculate the product of the numbers:

\(a \times b = 3003 \times 21 = 63063\)

Since the numbers are multiples of their HCF, we can express them as:

• \(a = 21m\)
• \(b = 21n\)

where \(m\) and \(n\) are co-prime (their HCF is 1).

Now, substitute these into the product equation:

\(21m \times 21n = 63063\)

which simplifies to:

\(441mn = 63063\)

Solving for \(mn\):

\(mn = \frac{63063}{441}\)

Calculate the division:

\(mn = 143\)

Now, we need to find two co-prime numbers whose product is 143. The factorization of 143 gives us:

\(143 = 11 \times 13\)

Thus, \(m = 11\) and \(n = 13\), since 11 and 13 are co-prime numbers.

The numbers are:

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

Finally, find the sum of these two numbers:

\(231 + 273 = 504\)

Thus, the sum of the two numbers is 504. Therefore, the correct answer is:

Option: 504

Answer 2

Let the two numbers be \( x \) and \( y \). From the property of LCM and HCF, we know:

\[ \text{LCM}(x, y) \times \text{HCF}(x, y) = x \times y \]

Substituting the given values:

\[ 3003 \times 21 = x \times y \]

\[ x \times y = 63063 \]

Let \( x = 21a \) and \( y = 21b \), where \( a \) and \( b \) are coprime. Hence:

\[ \text{LCM}(x, y) = 21 \times \text{LCM}(a, b) = 3003 \]

\[ \text{LCM}(a, b) = \frac{3003}{21} = 143 \]

Now, \( a \times b = \frac{x \times y}{21^2} = \frac{63063}{441} = 143 \).

Thus, the numbers \( a \) and \( b \) are the factors of 143, which are 11 and 13.

Therefore, \( x = 21 \times 11 = 231 \) and \( y = 21 \times 13 = 273 \). The sum of the numbers is:

\[ 231 + 273 = 504 \]