Finding the Smaller Number Using HCF and LCM Properties

This problem involves understanding the relationship between the Highest Common Factor (HCF), the Least Common Multiple (LCM), and the two numbers themselves. Let the two numbers be \(N_1\) and \(N_2\), and let their HCF be \(H\) and their LCM be \(L\).

We know a fundamental property connecting HCF and LCM with the two numbers:

\(N_1 \times N_2 = H \times L\)

Another important property relates the numbers, their HCF, and their LCM. If the two numbers are expressed as \(N_1 = H \times a\) and \(N_2 = H \times b\), where \(a\) and \(b\) are coprime (meaning their HCF is 1), then their LCM is given by:

\(L = H \times a \times b\)

In this problem, we are given:

• HCF (\(H\)) = 4
• The two other factors of LCM are 5 and 7.

The phrase "the two other factors of LCM are 5 and 7" implies that when the LCM is divided by the HCF, the result is a number whose factors are 5 and 7. In the context of \(L = H \times a \times b\), these "other factors" are typically the coprime parts \(a\) and \(b\). So, we can say that \(a=5\) and \(b=7\) (or vice versa).

Using the formula \(L = H \times a \times b\), we can calculate the LCM:

\(L = 4 \times 5 \times 7 = 140\)

Now, we need to find the two numbers, \(N_1\) and \(N_2\), such that their HCF is 4 and their LCM is 140. We know the numbers are of the form \(4a\) and \(4b\), where \(a\) and \(b\) are coprime and \(ab = L/H\).

Let's calculate \(ab\):

\(ab = \frac{L}{H} = \frac{140}{4} = 35\)

We need to find pairs of coprime integers \((a, b)\) such that their product \(ab = 35\). The pairs of factors of 35 are (1, 35) and (5, 7). Both pairs consist of coprime numbers:

• Pair 1: \(a=1, b=35\). HCF(1, 35) = 1 (coprime).
• Pair 2: \(a=5, b=7\). HCF(5, 7) = 1 (coprime).

Now we find the two numbers \(N_1 = 4a\) and \(N_2 = 4b\) for each pair:

• Using Pair 1 (\(a=1, b=35\)): The numbers are \(N_1 = 4 \times 1 = 4\) and \(N_2 = 4 \times 35 = 140\). The smaller number is 4.
• Using Pair 2 (\(a=5, b=7\)): The numbers are \(N_1 = 4 \times 5 = 20\) and \(N_2 = 4 \times 7 = 28\). The smaller number is 20.

We look at the given options: A. 10, B. 14, C. 20, D. 28.

Comparing the calculated smaller numbers (4 and 20) with the options, we see that 20 is one of the options (Option C).

Therefore, the two numbers are 20 and 28, and the smaller of the two numbers is 20.

The smaller number that matches one of the options is 20.

Revision Table: HCF, LCM, and Number Relationships

Additional Information on HCF and LCM Problems

Understanding the relationship between HCF and LCM is crucial for solving many problems in number theory and quantitative aptitude. The HCF is made up of the common prime factors raised to the lowest power, while the LCM is made up of all prime factors (common and uncommon) raised to the highest power found in either number.

Let the prime factorization of two numbers be:

\(N_1 = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}\)

\(N_2 = p_1^{b_1} p_2^{b_2} \dots p_k^{b_k}\)

(where \(p_i\) are prime numbers and \(a_i, b_i \ge 0\))

Then:

• HCF \(= p_1^{\min(a_1, b_1)} p_2^{\min(a_2, b_2)} \dots p_k^{\min(a_k, b_k)}\)
• LCM \(= p_1^{\max(a_1, b_1)} p_2^{\max(a_2, b_2)} \dots p_k^{\max(a_k, b_k)}\)

The relationship \(N_1 \times N_2 = \text{HCF} \times \text{LCM}\) can be verified using prime factorizations. \(a_i + b_i = \min(a_i, b_i) + \max(a_i, b_i)\). Summing over all prime factors, \(\sum (a_i + b_i) = \sum \min(a_i, b_i) + \sum \max(a_i, b_i)\), which translates to the product property.

In the context of this problem, the HCF (4) contributes factors of \(2^2\) to both numbers. The "other factors" (5 and 7) are the prime factors unique to each number after the common factor is removed. Thus, one number gets the factor 5, and the other gets the factor 7. Since 5 and 7 are coprime, this fits the structure \(N_1 = H \times a\) and \(N_2 = H \times b\) where \(a=5, b=7\) (or vice versa). This gives the numbers \(4 \times 5 = 20\) and \(4 \times 7 = 28\).