Finding the Sum of Reciprocals Using Sum, HCF, and LCM

This problem asks us to find the sum of the reciprocals of two numbers, given their sum, Highest Common Factor (HCF), and Least Common Multiple (LCM).

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

We are given the following information:

• Sum of the two numbers: \(a + b = 18\)
• HCF of the two numbers: HCF\((a, b) = 3\)
• LCM of the two numbers: LCM\((a, b) = 54\)

We need to find the sum of their reciprocals, which is \(\frac{1}{a} + \frac{1}{b}\).

Relationship Between Numbers, HCF, and LCM

A fundamental property of two positive integers is that the product of the numbers is equal to the product of their HCF and LCM.

Mathematically, this is expressed as:

\(a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)\)

Calculating the Product of the Numbers

Using the given values, we can calculate the product of the two numbers:

\(a \times b = 3 \times 54\)

\(a \times b = 162\)

Calculating the Sum of Reciprocals

The sum of the reciprocals of the two numbers \(a\) and \(b\) is \(\frac{1}{a} + \frac{1}{b}\). To add these fractions, we find a common denominator, which is the product \(a \times b\).

\(\frac{1}{a} + \frac{1}{b} = \frac{b}{a \times b} + \frac{a}{a \times b}\)

\(\frac{1}{a} + \frac{1}{b} = \frac{a + b}{a \times b}\)

Substituting Values and Finding the Result

We know the sum (\(a + b\)) is 18 and the product (\(a \times b\)) is 162. We can substitute these values into the expression for the sum of reciprocals:

Sum of reciprocals \(= \frac{18}{162}\)

Simplifying the Fraction

Now, we simplify the fraction \(\frac{18}{162}\). Both the numerator and the denominator are divisible by 18.

• \(18 \div 18 = 1\)
• \(162 \div 18 = 9\) (since \(18 \times 9 = 162\))

So, the simplified fraction is \(\frac{1}{9}\).

Therefore, the sum of the reciprocals of the two numbers is \(\frac{1}{9}\).

Revision Table: Key Concepts

Additional Information: Understanding HCF and LCM Problems

Problems involving HCF and LCM often test your understanding of their definitions and the relationship \(a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)\). While this property holds for any two positive integers, finding the specific numbers \(a\) and \(b\) might require additional conditions, such as their sum or difference.

In this problem, even if finding the specific integer values of \(a\) and \(b\) that satisfy all three conditions (sum=18, HCF=3, LCM=54) was complex or impossible for positive integers (as 3x+3y=18 implies x+y=6, and 9xy=162 implies xy=18, but no integer pair (x,y) exists such that x+y=6 and xy=18, simultaneously coprime), the question about the sum of reciprocals relies only on the values of \(a+b\) and \(ab\). We were given \(a+b\) directly, and we could find \(ab\) using the HCF and LCM property. Therefore, we could calculate the sum of reciprocals \(\frac{a+b}{ab}\) without needing to find the individual numbers themselves.

This highlights that sometimes, to answer a question about numbers, you only need their sum and product, not the numbers individually.