Understanding the Problem

The goal is to find two specific numbers. We are given two pieces of information about these numbers: their sum is 80, and their Least Common Multiple (LCM) is 558.

• Given Sum: 80
• Given L.C.M.: 558

Key Concepts: Sum, LCM, and HCF

Let the two numbers be represented by '$a$' and '$b$'.

We can use the following mathematical properties:

1. Sum Property: The sum of the two numbers is given: $a + b = 80$
2. Product Property: The product of two numbers is equal to the product of their Highest Common Factor (HCF) and their Least Common Multiple (LCM). $a \times b = HCF(a, b) \times LCM(a, b)$

Let $HCF(a, b)$ be represented by '$h$'. A useful way to represent the numbers is by using their HCF. If $h$ is the HCF of $a$ and $b$, we can write:

• $a = hx$
• $b = hy$

Here, '$x$' and '$y$' must be coprime numbers, meaning their HCF is 1 ($HCF(x, y) = 1$).

Calculating the HCF

Let's apply these representations to the given information:

• Sum: $hx + hy = 80 \implies h(x+y) = 80$
• LCM: The LCM of $hx$ and $hy$ is $hxy$. So, $hxy = 558$.

From the equation $h(x+y) = 80$, we know that $h$ must be a divisor of 80.

From the equation $hxy = 558$, we know that $h$ must be a divisor of 558.

Therefore, $h$ must be a common divisor of both 80 and 558. To find the specific value of $h$, we can find the Highest Common Factor (HCF) of 80 and 558.

Let's find the prime factorization of each number:

• $80 = 8 \times 10 = (2 \times 2 \times 2) \times (2 \times 5) = 2^4 \times 5$
• $558 = 2 \times 279 = 2 \times (9 \times 31) = 2 \times (3 \times 3) \times 31 = 2 \times 3^2 \times 31$

The common prime factor is 2, raised to the lowest power (which is $2^1$). So, the HCF of 80 and 558 is 2.

This means the HCF of our two numbers is $h=2$.

Finding the Coprime Factors

Now that we know $h=2$, we can find the sum ($x+y$) and product ($xy$) of the coprime factors $x$ and $y$.

1. Using the sum equation: $h(x+y) = 80 \implies 2(x+y) = 80$. Dividing by 2, we get $x+y = 40$.
2. Using the LCM equation: $hxy = 558 \implies 2xy = 558$. Dividing by 2, we get $xy = 279$.

Our task now is to find two numbers ($x$ and $y$) that satisfy these two conditions: their sum is 40, and their product is 279. Remember, $x$ and $y$ must also be coprime.

Solving for the Numbers

We can find the numbers $x$ and $y$ by solving the system of equations:

• $x+y = 40$
• $xy = 279$

We can use a quadratic equation approach. If $x$ and $y$ are the roots of a quadratic equation, the equation can be written as $t^2 - (sum \, of \, roots)t + (product \, of \, roots) = 0$.

Substituting our values: $t^2 - (40)t + 279 = 0$

We solve this quadratic equation using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $a=1$, $b=-40$, and $c=279$.

Calculating the discriminant ($D = b^2 - 4ac$): $D = (-40)^2 - 4(1)(279)$ $D = 1600 - 1116$ $D = 484$

Now, find the square root of the discriminant: $\sqrt{D} = \sqrt{484} = 22$

Now, apply the formula for $t$: $t = \frac{-(-40) \pm 22}{2(1)}$ $t = \frac{40 \pm 22}{2}$

This gives us two possible values for $t$, which are our numbers $x$ and $y$:

• $t_1 = \frac{40 + 22}{2} = \frac{62}{2} = 31$
• $t_2 = \frac{40 - 22}{2} = \frac{18}{2} = 9$

So, the coprime factors are $x=9$ and $y=31$ (or vice versa). Let's quickly check if they are coprime: HCF(9, 31) = 1. Yes, they are.

Now we find the original numbers $a$ and $b$ using $a=hx$ and $b=hy$:

• $a = 2 \times 9 = 18$
• $b = 2 \times 31 = 62$

The two numbers are 18 and 62.

Verification of the Numbers

Let's check if the numbers 18 and 62 satisfy the original conditions:

• Check Sum: $18 + 62 = 80$. This matches the given sum.
• Check LCM: First, find the HCF of 18 and 62. $18 = 2 \times 3^2$ $62 = 2 \times 31$ $HCF(18, 62) = 2$. Now, use the product property: $LCM(18, 62) = \frac{18 \times 62}{HCF(18, 62)} = \frac{18 \times 62}{2}$. $LCM(18, 62) = 9 \times 62 = 558$. This matches the given LCM.

Both conditions are met, confirming our answer.

Checking the Options Provided

We can also verify the options given in the question to see which pair satisfies both conditions:

By checking the options, we confirm that the pair (18, 62) is the correct answer as it satisfies both the sum and LCM requirements.

Final Answer

The two numbers are 18 and 62.