Finding the Numbers: Problem Overview

We are asked to find two numbers. We know two key pieces of information about these numbers:

• Their sum is 57.
• Their Least Common Multiple (LCM) is 812.

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

From the problem statement, we have:

• Equation 1: \( a + b = 57 \)
• Equation 2: \( \text{LCM}(a, b) = 812 \)

Checking Provided Options Systematically

We can test each option pair to see if they satisfy both the sum and LCM conditions.

The pair 28, 29 is the only one that satisfies both the sum and LCM conditions.

Mathematical Solution Explained

We can use the relationship between two numbers (a, b), their Greatest Common Divisor (GCD), and their Least Common Multiple (LCM):

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

Let \( g = \text{GCD}(a, b) \). We can express the numbers as \( a = gx \) and \( b = gy \), where \( x \) and \( y \) are coprime integers (their GCD is 1).

Using the sum condition:

\( a + b = 57 \)

\( gx + gy = 57 \)

\( g(x + y) = 57 \)

This means \( g \) must be a divisor of 57. The divisors of 57 are 1, 3, 19, 57.

Using the LCM condition:

\( \text{LCM}(a, b) = 812 \)

Since \( a = gx \) and \( b = gy \) and \( x, y \) are coprime, \( \text{LCM}(a, b) = gxy \).

\( gxy = 812 \)

This means \( g \) must also be a divisor of 812.

We need a common divisor of 57 and 812. Let's list the divisors:

• Divisors of 57: {1, 3, 19, 57}
• Divisors of 812: {1, 2, 4, 7, 14, 28, 29, 58, 116, 203, 406, 812}

The only common divisor is 1. Thus, \( g = \text{GCD}(a, b) = 1 \).

Now we update our equations with \( g = 1 \):

1. \( 1(x + y) = 57 \) => \( x + y = 57 \)
2. \( 1xy = 812 \) => \( xy = 812 \)

We need two coprime numbers \( x \) and \( y \) that add up to 57 and multiply to 812.

We can think of \( x \) and \( y \) as the roots of a quadratic equation \( t^2 - (\text{sum of roots})t + (\text{product of roots}) = 0 \).

\( t^2 - (x+y)t + xy = 0 \)

\( t^2 - 57t + 812 = 0 \)

Solve for \( t \) using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\( t = \frac{57 \pm \sqrt{(-57)^2 - 4(1)(812)}}{2(1)} \)

\( t = \frac{57 \pm \sqrt{3249 - 3248}}{2} \)

\( t = \frac{57 \pm \sqrt{1}}{2} \)

\( t = \frac{57 \pm 1}{2} \)

This gives two possible values for \( t \):

• \( t_1 = \frac{57 + 1}{2} = \frac{58}{2} = 29 \)
• \( t_2 = \frac{57 - 1}{2} = \frac{56}{2} = 28 \)

So, the coprime numbers \( x \) and \( y \) are 28 and 29.

Since \( g = 1 \), the original numbers \( a \) and \( b \) are:

\( a = g \times x = 1 \times 28 = 28 \)

\( b = g \times y = 1 \times 29 = 29 \)

The two numbers are 28 and 29.