If 20/2^(5) = {1}{2^(m)} + {1}{2^(n)} what is nm ?

1. Translate the problem requirements: We need to express \(\frac{20}{2^5}\) as the sum of two unit fractions where each denominator is a power of 2 (specifically \(2^m\) and \(2^n\)), then find the product nm.
2. Simplify the left side to identify the target: Convert \(\frac{20}{2^5}\) into its simplest form to see what we're working toward.
3. Find compatible unit fractions: Determine which two unit fractions with power-of-2 denominators can sum to our target fraction by using common denominator logic.
4. Extract exponents and calculate the product: Once we identify the unit fractions, determine the values of m and n, then compute nm.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for in everyday terms. We have a fraction on the left side: \(\frac{20}{2^5}\). On the right side, we need to find two "unit fractions" (fractions with 1 in the numerator) where the bottom parts are powers of 2.

Think of it like this: imagine you have a pizza cut into \(2^5\) pieces, and you take 20 of those pieces. Now you want to represent that same amount as two separate portions, where each portion is exactly one piece from pizzas that are cut into \(2^m\) pieces and \(2^n\) pieces respectively.

Our goal is to find the values m and n, then multiply them together to get nm.

Process Skill: TRANSLATE - Converting the equation format into a clear mathematical goal

2. Simplify the left side to identify the target

Let's work out what \(\frac{20}{2^5}\) actually equals in simpler terms.

First, \(2^5 = 32\), so we have \(\frac{20}{32}\).

Now let's simplify this fraction by finding the greatest common factor of 20 and 32:

• 20 = 4 × 5
• 32 = 4 × 8

So \(\frac{20}{32} = \frac{4 \times 5}{4 \times 8} = \frac{5}{8}\)

Since \(8 = 2^3\), we can write this as: \(\frac{5}{2^3}\)

So our target is: \(\frac{5}{2^3} = \frac{1}{2^m} + \frac{1}{2^n}\)

3. Find compatible unit fractions

Now we need to express \(\frac{5}{2^3}\) as the sum of two unit fractions with power-of-2 denominators.

Let's think about this step by step. We have \(\frac{5}{8}\), and we want to write it as \(\frac{1}{2^m} + \frac{1}{2^n}\).

To add fractions, we need a common denominator. The most natural common denominator would be the larger of \(2^m\) and \(2^n\). Let's assume n ≥ m, so our common denominator is \(2^n\).

Then: \(\frac{1}{2^m} + \frac{1}{2^n} = \frac{2^{(n-m)}}{2^n} + \frac{1}{2^n} = \frac{2^{(n-m)} + 1}{2^n}\)

For this to equal \(\frac{5}{2^3}\), we need:

• \(2^n = 2^3 = 8\), so n = 3
• \(2^{(n-m)} + 1 = 5\), so \(2^{(3-m)} + 1 = 5\), which means \(2^{(3-m)} = 4 = 2^2\)
• Therefore 3-m = 2, so m = 1

Let's verify: \(\frac{1}{2^1} + \frac{1}{2^3} = \frac{1}{2} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8}\) ✓

Process Skill: INFER - Drawing the non-obvious conclusion about the relationship between exponents

4. Extract exponents and calculate the product

From our work above, we found:

• m = 1
• n = 3

Therefore: \(nm = 1 \times 3 = 3\)

Final Answer

The product nm = 3, which corresponds to answer choice (B).

We can double-check our work: \(\frac{1}{2^1} + \frac{1}{2^3} = \frac{1}{2} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8} = \frac{20}{32} = \frac{20}{2^5}\) ✓

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Students may try to solve for m and n independently without recognizing that the equation requires finding a systematic relationship between the two variables. They might attempt to guess values or use trial-and-error without establishing the constraint that both denominators must be powers of 2.

Faltering Point 2: Students often skip the crucial step of simplifying \(\frac{20}{2^5}\) to its reduced form \(\frac{5}{8}\). Instead, they work directly with the unsimplified fraction, making the algebra much more complex and leading to incorrect setup of equations.

Faltering Point 3: Students may not recognize that they need to express the sum of unit fractions using a common denominator approach. They might try to work backwards from the answer choices or use other inefficient methods instead of systematically setting up the equation \(\frac{2^{(n-m)} + 1}{2^n} = \frac{5}{2^3}\).

Errors while executing the approach

Faltering Point 1: When setting up the equation \(2^{(n-m)} + 1 = 5\), students frequently make arithmetic errors in solving \(2^{(n-m)} = 4\). They may incorrectly conclude that n-m = 3 instead of n-m = 2, or make sign errors when solving for m given that n = 3.

Faltering Point 2: Students often make errors when adding fractions with different denominators. For example, when verifying \(\frac{1}{2} + \frac{1}{8}\), they might incorrectly add the denominators (getting \(\frac{1}{10}\)) or make mistakes in finding the common denominator of 8.

Faltering Point 3: Students may assume the wrong ordering of m and n (assuming m ≥ n instead of n ≥ m), which leads to setting up the common denominator incorrectly and arriving at the wrong values for the exponents.

Errors while selecting the answer

Faltering Point 1: Students might calculate m + n = 1 + 3 = 4 instead of the required product nm = 1 × 3 = 3. This is a common error where students confuse addition with multiplication when combining the final values.