If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different...

1. Translate the problem requirements: We need to find which fraction produces the repeating decimal with the longest sequence of different digits before the pattern repeats (this is called the period length).
2. Identify the key insight about period length: For a fraction in lowest terms, the period length of its decimal expansion depends on the denominator - specifically on the smallest positive integer that makes the denominator divide a number of the form \(10^n - 1\).
3. Convert fractions to have comparable denominators: Ensure all fractions are in lowest terms and identify the true denominators that determine period length.
4. Calculate period lengths by testing powers of 10: For each denominator, find the smallest n where \(10^n \equiv 1 \pmod{\mathrm{denominator}}\), which gives us the period length.
5. Compare results to identify the maximum: The fraction with the largest period length will have the longest sequence of different digits.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what we're looking for. When we convert a fraction to a decimal, we often get repeating decimals. For example, \(\frac{1}{3} = 0.333...\) where "3" repeats forever. The question asks for the fraction that has the "longest sequence of different digits" before the pattern repeats.

This sequence of different digits before repetition is called the period length. For instance:

• \(\frac{1}{3} = 0.333...\) has period length 1 (only "3" repeats)
• \(\frac{1}{7} = 0.142857142857...\) has period length 6 ("142857" repeats)

We need to find which of our five fractions has the longest period length.

Process Skill: TRANSLATE - Converting the phrase "longest sequence of different digits" to the mathematical concept of period length

2. Identify the key insight about period length

Here's the key insight: The period length of a fraction's decimal expansion depends on its denominator (when the fraction is in lowest terms).

Specifically, we need to find the smallest positive number n such that when we multiply the denominator by some whole number, we get a result that ends in 999...9 (n nines). In mathematical terms, we're looking for the smallest n where \(10^n - 1\) is divisible by the denominator.

For example, with denominator 3: \(10^1 - 1 = 9\), and \(9 \div 3 = 3\) exactly. So the period length is 1.

3. Convert fractions to have comparable denominators

Let's check if our fractions are in lowest terms and identify the denominators that will determine period length:

• A) \(\frac{2}{11}\): Already in lowest terms, denominator = 11
• B) \(\frac{1}{3}\): Already in lowest terms, denominator = 3
• C) \(\frac{41}{99}\): Let's check if this can be simplified. Since \(99 = 9 \times 11 = 3^2 \times 11\), and 41 is prime (not divisible by 3 or 11), this is in lowest terms. Denominator = 99
• D) \(\frac{2}{3}\): Already in lowest terms, denominator = 3
• E) \(\frac{23}{37}\): Since 37 is prime and doesn't divide 23, this is in lowest terms. Denominator = 37

So we need to find period lengths for denominators: 11, 3, 99, 3, and 37.

4. Calculate period lengths by testing powers of 10

For each denominator, we'll find the smallest n where \(10^n \equiv 1 \pmod{\mathrm{denominator}}\), which means \(10^n - 1\) is divisible by the denominator.

For denominator 3:

\(10^1 - 1 = 9\), and \(9 \div 3 = 3\) ✓

Period length = 1

For denominator 11:

\(10^1 - 1 = 9\), and \(9 \div 11 = 0\) remainder 9 ✗

\(10^2 - 1 = 99\), and \(99 \div 11 = 9\) ✓

Period length = 2

For denominator 37:

We need to test: \(10^1 - 1 = 9\) (not divisible by 37)

\(10^2 - 1 = 99\) (not divisible by 37)

\(10^3 - 1 = 999\) (\(999 \div 37 = 27\), so this works ✓)

Period length = 3

For denominator 99:

Since \(99 = 9 \times 11\), and we know period length for 11 is 2, we can use the fact that for 99:

\(10^2 - 1 = 99\), and \(99 \div 99 = 1\) ✓

Period length = 2

Process Skill: INFER - Using the relationship between 99 and 11 to simplify our calculation

5. Compare results to identify the maximum

Let's summarize our period lengths:

• A) \(\frac{2}{11}\): Period length = 2
• B) \(\frac{1}{3}\): Period length = 1
• C) \(\frac{41}{99}\): Period length = 2
• D) \(\frac{2}{3}\): Period length = 1
• E) \(\frac{23}{37}\): Period length = 3

The longest period length is 3, which corresponds to choice E) \(\frac{23}{37}\).

Final Answer

The fraction \(\frac{23}{37}\) will have the longest sequence of different digits when written as a repeating decimal, with a period length of 3.

Answer: E

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "longest sequence of different digits"

Students often confuse this phrase with "most number of digits after decimal point" or "largest decimal value." They might try to convert each fraction to decimal form and count total digits, missing that we need the length of the repeating pattern before it cycles back.

2. Focusing on numerator instead of denominator

Students may incorrectly assume that larger numerators lead to longer decimal expansions. They might rank fractions like \(\frac{41}{99}\) or \(\frac{23}{37}\) as having longer periods simply because 41 and 23 are larger numbers, not realizing that period length is determined primarily by the denominator.

3. Not checking if fractions are in lowest terms

Students might work directly with the given fractions without reducing them first. For example, they might analyze \(\frac{41}{99}\) without recognizing it's already in lowest terms, or fail to realize that period length calculations require working with the reduced form of the fraction.

Errors while executing the approach

1. Arithmetic errors in testing powers of 10

Students frequently make calculation mistakes when testing whether \(10^n - 1\) is divisible by each denominator. For example, they might incorrectly calculate \(10^3 - 1 = 999 \div 37\), or make division errors that lead them to wrong period lengths.

2. Stopping too early in the period length calculation

For denominators like 37, students might test only \(10^1 - 1 = 9\) and \(10^2 - 1 = 99\), find these aren't divisible by 37, and incorrectly conclude the period length is very long or give up, rather than continuing to test \(10^3 - 1 = 999\).

3. Misunderstanding the divisibility test

Students might confuse the condition they're testing. Instead of checking if \(10^n - 1\) is divisible by the denominator, they might test if \(10^n\) is divisible by the denominator, or check divisibility in the wrong direction, leading to incorrect period length calculations.

Errors while selecting the answer

1. Choosing based on denominator size rather than period length

After calculating period lengths, students might still default to choosing the fraction with the largest denominator (like \(\frac{41}{99}\)) assuming bigger denominators automatically mean longer periods, rather than selecting based on their calculated period length results.