If a, b and c are positive integers such that \(\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{2}}}} =\dfrac{16}{23}\) , then what is the mean of a, b and c?
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Understanding the Problem: Continued Fractions and Mean
The problem asks us to find the mean of three positive integers, a, b, and c. These integers are related by a given equation involving a continued fraction. The equation is:
\(\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{2}}}} =\dfrac{16}{23}\)
To find the mean of a, b, and c, we first need to determine their individual values by solving this equation. The mean is calculated as \(\dfrac{a+b+c}{3}\).
Solving the Continued Fraction Equation
We can solve the equation by successively taking the reciprocal of both sides and expressing the result as a mixed number. This process helps us identify the integer parts, which correspond to a, b, and c.
Step 1: Isolate 'a'
Given the equation:
\(\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{2}}}} =\dfrac{16}{23}\)
Take the reciprocal of both sides:
\(a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{2}}} = \dfrac{23}{16}\)
Express the right side as a mixed number:
\(\dfrac{23}{16} = 1 + \dfrac{7}{16}\)
Comparing this with \(a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{2}}}\), since a is a positive integer, we can identify the integer part:
\(a = 1\)
This leaves us with the remaining fractional part equation:
\(\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{2}}} = \dfrac{7}{16}\)
Step 2: Isolate 'b'
Take the reciprocal of the equation from Step 1:
\(b+\dfrac{1}{c+\dfrac{1}{2}} = \dfrac{16}{7}\)
Express the right side as a mixed number:
\(\dfrac{16}{7} = 2 + \dfrac{2}{7}\)
Comparing this with \(b+\dfrac{1}{c+\dfrac{1}{2}}\), since b is a positive integer, we identify:
\(b = 2\)
This leaves us with:
\(\dfrac{1}{c+\dfrac{1}{2}} = \dfrac{2}{7}\)
Step 3: Isolate 'c'
Take the reciprocal of the equation from Step 2:
\(c+\dfrac{1}{2} = \dfrac{7}{2}\)
We can write \(\dfrac{7}{2}\) as \(3.5\) or \(3 + \dfrac{1}{2}\).
\(c+\dfrac{1}{2} = 3+\dfrac{1}{2}\)
Subtracting \(\dfrac{1}{2}\) from both sides gives:
\(c = 3\)
So, we have found the positive integer values: a = 1, b = 2, and c = 3.
Calculating the Mean of a, b, and c
Now that we have the values of a, b, and c, we can calculate their mean using the formula:
Mean = \(\dfrac{a+b+c}{3}\)
Substitute the values \(a=1\), \(b=2\), and \(c=3\):
Mean = \(\dfrac{1+2+3}{3}\)
Mean = \(\dfrac{6}{3}\)
Mean = \(2\)
The mean of a, b, and c is 2.
Revision Table: Key Concepts
| Concept | Description | Formula/Example |
|---|---|---|
| Continued Fraction | An expression obtained through an iterative process of representing a number as a sum of its integer part and the reciprocal of another number, and so on. | e.g., \(a_0 + \dfrac{1}{a_1 + \dfrac{1}{a_2 + \dots}}\) |
| Mean | The average of a set of numbers. It is the sum of the numbers divided by the count of the numbers. | Mean = \(\dfrac{\text{Sum of values}}{\text{Number of values}}\) |
| Reciprocal | The reciprocal of a number x is 1/x. Taking the reciprocal flips the numerator and the denominator of a fraction. | Reciprocal of \(\dfrac{p}{q}\) is \(\dfrac{q}{p}\) |
| Positive Integer | A whole number greater than 0 (1, 2, 3, ...). | 1, 2, 3, 4, 5, ... |
Additional Information: Properties of Continued Fractions
Continued fractions are a way to represent real numbers. They have interesting properties, especially when representing rational and irrational numbers.
- Finite continued fractions, like the one in the problem which terminates with 1/2, represent rational numbers.
- Infinite continued fractions represent irrational numbers. For example, the golden ratio \(\phi\) can be represented as an infinite continued fraction consisting only of ones: \(1 + \dfrac{1}{1 + \dfrac{1}{1 + \dots}}\).
- Solving continued fraction equations like this involves using the property that if two continued fractions are equal, their corresponding integer and fractional parts at each step must be equal, provided the expansion is unique (which is often the case for a specific form like the one in the problem).
- This method of equating integer and fractional parts is similar to the Euclidean algorithm used for finding the greatest common divisor of two numbers.
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